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 Exercise 5.5

Differentiate the functions given in Exercises 1 to 11 w.r.t. x.

1. cos x. cos 2x. cos 3x

Solun:- Let y = cos x. cos 2x. cos 3x

Differentiate w.r.t. x:-

We know that:-

{"type":"$","code":"$2.\\,{\\sqrt[]{\\frac{\\left(x-1\\right)\\left(x-2\\right)}{\\left(x-3\\right)\\left(x-4\\right)\\left(x-5\\right)}}}$","font":{"color":"#434343","family":"Arial","size":14},"id":"6","ts":1598260281133,"cs":"Z8kmxE7690f4+GZ8Cpnjkg==","size":{"width":180,"height":41}}

Solun:- Let

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Taking log both side:-

{"id":"10","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\log_{}\\left(y\\right)}&={\\log_{}\\left({\\sqrt[]{\\frac{\\left(x-1\\right)\\left(x-2\\right)}{\\left(x-3\\right)\\left(x-4\\right)\\left(x-5\\right)}}}\\right)}\\\\\n{\\log_{}\\left(y\\right)}&={\\frac{1}{2}\\log_{}\\left(\\frac{\\left(x-1\\right)\\left(x-2\\right)}{\\left(x-3\\right)\\left(x-4\\right)\\left(x-5\\right)}\\right)}\t\n\\end{align*}","type":"align*","ts":1598262824019,"cs":"MEaV/z+Zhzgt2g1gM/ndXw==","size":{"width":280,"height":96}}{"font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"11","code":"\\begin{align*}\n{\\log_{}\\left(y\\right)}&={\\frac{1}{2}\\left[\\log_{}\\left(x-1\\right)+\\log_{}\\left(x-2\\right)-\\log_{}\\left(x-3\\right)-\\log_{}\\left(x-4\\right)-\\log_{}\\left(x-5\\right)\\right]}\t\n\\end{align*}","ts":1598262784412,"cs":"098a9NQ76MWCKxDA2YWfTA==","size":{"width":512,"height":32}}

Differentiate w.r.t. x:-

{"font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"12","code":"\\begin{align*}\n{\\frac{1}{y}\\diff{y}{x}}&={\\frac{1}{2}\\left[\\frac{1}{x-1}+\\frac{1}{x-2}-\\frac{1}{x-3}-\\frac{1}{x-4}-\\frac{1}{x-5}\\right]}\\\\\n{\\diff{y}{x}}&={\\frac{y}{2}\\left[\\frac{1}{x-1}+\\frac{1}{x-2}-\\frac{1}{x-3}-\\frac{1}{x-4}-\\frac{1}{x-5}\\right]}\\\\\n{\\diff{y}{x}}&={\\frac{1}{2}{\\sqrt[]{\\frac{\\left(x-1\\right)\\left(x-2\\right)}{\\left(x-3\\right)\\left(x-4\\right)\\left(x-5\\right)}}}\\left[\\frac{1}{x-1}+\\frac{1}{x-2}-\\frac{1}{x-3}-\\frac{1}{x-4}-\\frac{1}{x-5}\\right]}\t\n\\end{align*}","ts":1598263084190,"cs":"05AFkkIL6UXeCkyjvaWqvA==","size":{"width":556,"height":133}}

3. (log x)cos x

Solun:- Let y = (log x)cos x

Taking log both side:-

log y = [log((log x)cos x)]

We know that log xn = n.(log x)

log y = cos x.[log(log x)]

Differentiate w.r.t. x:-

{"type":"$","code":"$\\diff{\\left(\\log_{}y\\right)}{x}=\\diff{\\left[\\cos x\\times\\left\\{\\log_{}\\left(\\log_{}x\\right)\\right\\}\\right]}{x}$","font":{"family":"Arial","size":12,"color":"#000000"},"id":"13","ts":1598263495468,"cs":"iBsB6FKJ+D1S633qa/G6zw==","size":{"width":224,"height":28}}

We know that:-

{"type":"$","code":"$\\diff{\\left(u.v\\right)}{x}=u.\\diff{\\left(v\\right)}{x}+v\\diff{\\left(u\\right)}{x}$","font":{"color":"#000000","family":"Arial","size":10},"id":"15","ts":1598264093356,"cs":"f3YXe902kC2/qgYXzFOv0g==","size":{"width":156,"height":21}}

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"14","type":"align*","code":"\\begin{align*}\n{\\frac{1}{y}\\diff{y}{x}}&={\\cos x\\times \\diff{\\left[\\log_{}\\left(\\log_{}x\\right)\\right]}{x}+\\log_{}\\left(\\log_{}x\\right)\\diff{\\left(\\cos x\\right)}{x}}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\cos x\\times\\frac{1}{\\log_{}x}\\times\\frac{1}{x}+\\log_{}\\left(\\log_{}x\\right)\\left(-\\sin x\\right)}\\\\\n{\\diff{y}{x}}&={y\\left[\\frac{\\cos x}{x\\log_{}x}-\\sin x\\log_{}\\left(\\log_{}x\\right)\\right]}\\\\\n{\\diff{y}{x}}&={\\left(\\log_{}x\\right)^{\\cos x}\\left[\\frac{\\cos x}{x\\log_{}x}-\\sin x\\log_{}\\left(\\log_{}x\\right)\\right]}\t\n\\end{align*}","ts":1598264009456,"cs":"F2WJY1AOLjrQFJLIR70Ohg==","size":{"width":349,"height":164}}

4. xx - 2sin x

Solun:- Let y = xx - 2sin x

Let u = xx  and v = 2sin x

⇒ y = u - v

⇒ u = xx

Taking log both side:-

⇒ log u = log xx

We know that log xn = n.(log x)

⇒ log u = x.log x

Differentiate w.r.t. x:-

{"code":"\\begin{align*}\n{\\frac{1}{u}\\diff{u}{x}}&={x\\diff{\\left(\\log_{}x\\right)}{x}+\\left(\\log_{}x\\right)\\diff{x}{x}}\\\\\n{\\diff{u}{x}}&={u\\left[x\\times\\frac{1}{x}+\\left(\\log_{}x\\right)\\right]}\\\\\n{\\diff{u}{x}}&={x^{x}\\left[1+\\left(\\log_{}x\\right)\\right]}\t\n\\end{align*}","type":"align*","font":{"color":"#000000","family":"Arial","size":10},"id":"16","ts":1598264937647,"cs":"dE1GNwAl34FeXryHTTvpCQ==","size":{"width":220,"height":113}}

⇒ v = 2sin x

Taking log both side:-

⇒ log v = log 2sin x

We know that log xn = n.(log x)

⇒ log v = (sin x).log 2

Differentiate w.r.t. x:-

{"type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"id":"16","code":"\\begin{align*}\n{\\frac{1}{v}\\diff{v}{x}}&={\\sin x\\diff{\\left(\\log_{}2\\right)}{x}+\\left(\\log_{}2\\right)\\diff{\\left(\\sin x\\right)}{x}}\\\\\n{\\diff{v}{x}}&={v\\left[\\sin x\\times0+\\left(\\log_{}2\\right)\\times \\cos x\\right]}\\\\\n{\\diff{v}{x}}&={2^{\\sin x}\\left[0+\\left(\\log_{}2\\right)\\times \\cos x\\right]}\\\\\n{\\diff{v}{x}}&={2^{\\sin x}\\times \\cos x\\times \\log_{}2}\t\n\\end{align*}","ts":1598265504541,"cs":"VxKGjBTPr63A2bCdIb9unA==","size":{"width":273,"height":146}}

Then dy/dx = du/dx - dv/dx

⇒ dy/dx = xx(1+log x) - 2sin x.cos x.log 2

5. (x+3)2.(x+4)3.(x+5)4

Solun:- Let y = (x+3)2.(x+4)3.(x+5)4

Differentiate w.r.t. x:-

{"font":{"size":8,"family":"Arial","color":"#000000"},"type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\left(x+3\\right)^{2}\\left(x+4\\right)^{3}\\diff{\\left[\\left(x+5\\right)^{4}\\right]}{x}+\\left(x+4\\right)^{3}\\left(x+5\\right)^{4}\\diff{\\left[\\left(x+3\\right)^{2}\\right]}{x}+\\left(x+3\\right)^{2}\\left(x+5\\right)^{4}\\diff{\\left[\\left(x+4\\right)^{3}\\right]}{x}}\\\\\n{\\diff{y}{x}}&={4\\left(x+5\\right)^{3}\\left(x+3\\right)^{2}\\left(x+4\\right)^{3}+2\\left(x+3\\right)\\left(x+4\\right)^{3}\\left(x+5\\right)^{4}+3\\left(x+4\\right)^{2}\\left(x+3\\right)^{2}\\left(x+5\\right)^{4}}\\\\\n{\\diff{y}{x}}&={\\left(x+3\\right)\\left(x+4\\right)^{2}\\left(x+5\\right)^{3}\\left[4\\left(x+3\\right)\\left(x+4\\right)+2\\left(x+4\\right)\\left(x+5\\right)+3\\left(x+3\\right)\\left(x+5\\right)\\right]}\\\\\n{\\diff{y}{x}}&={\\left(x+3\\right)\\left(x+4\\right)^{2}\\left(x+5\\right)^{3}\\left[4\\left(x^{2}+7x+12\\right)+2\\left(x^{2}+9x+20\\right)+3\\left(x^{2}+8x+15\\right)\\right]}\\\\\n{\\diff{y}{x}}&={\\left(x+3\\right)\\left(x+4\\right)^{2}\\left(x+5\\right)^{3}\\left[9x^{2}+70x+133\\right]}\t\n\\end{align*}","id":"17","ts":1598531866865,"cs":"2uTQVyZo4UBgcEcImRBppw==","size":{"width":576,"height":168}}

{"id":"18","font":{"size":12,"color":"#434343","family":"Arial"},"type":"$","code":"$6.\\,\\left(x+\\frac{1}{x}\\right)^{x}+x^{\\left(x+\\frac{1}{x}\\right)}$","ts":1598268476637,"cs":"l2Qr139kOps9x3il7I1RGQ==","size":{"width":182,"height":26}}

Solun:- Let 

{"code":"$y=\\left(x+\\frac{1}{x}\\right)^{x}+x^{\\left(x+\\frac{1}{x}\\right)}$","id":"18","font":{"color":"#434343","family":"Arial","size":12},"type":"$","ts":1598268526742,"cs":"fBglypceQkp0rJk5O5bLAw==","size":{"width":200,"height":26}}

⇒ y = u + v

⇒ u = (x+1/x)x

Taking log both sides:-

⇒ log u = log (x+1/x)x

We know that log xn = n.(log x)

⇒ log u = x.log (x+1/x)

Differentiate w.r.t. x:-

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"19-0-0","code":"\\begin{align*}\n{\\frac{1}{u}\\diff{u}{x}}&={\\diff{\\left[x.\\log_{}\\left(x+\\frac{1}{x}\\right)\\right]}{x}}\\\\\n{\\frac{1}{u}\\frac{du}{dx}}&={x.\\diff{\\left[\\log_{}\\left(x+\\frac{1}{x}\\right)\\right]}{x}+\\log_{}\\left(x+\\frac{1}{x}\\right)\\diff{x}{x}}\\\\\n{\\frac{du}{dx}}&={u\\left[x.\\diff{\\left[\\log_{}\\left(x+\\frac{1}{x}\\right)\\right]}{x}+\\log_{}\\left(x+\\frac{1}{x}\\right)\\diff{x}{x}\\right]}\\\\\n{\\frac{du}{dx}}&={u\\left[x\\times\\frac{1}{\\left(x+\\frac{1}{x}\\right)}\\times\\left(1-\\frac{1}{x^{2}}\\right)+\\log_{}\\left(x+\\frac{1}{x}\\right)\\right]}\\\\\n{\\frac{du}{dx}}&={u\\left[x\\times\\frac{1}{\\frac{x^{2+1}}{x}}\\times\\left(\\frac{x^{2}-1}{x^{2}}\\right)+\\log_{}\\left(x+\\frac{1}{x}\\right)\\right]}\t\n\\end{align*}","ts":1598269471701,"cs":"Ip0gGLhrW+eNufAeIr51UA==","size":{"width":381,"height":240}}

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"19-0-0","code":"\\begin{align*}\n{\\frac{1}{u}\\diff{u}{x}}&={\\diff{\\left[x.\\log_{}\\left(x+\\frac{1}{x}\\right)\\right]}{x}}\\\\\n{\\frac{1}{u}\\frac{du}{dx}}&={x.\\diff{\\left[\\log_{}\\left(x+\\frac{1}{x}\\right)\\right]}{x}+\\log_{}\\left(x+\\frac{1}{x}\\right)\\diff{x}{x}}\\\\\n{\\frac{du}{dx}}&={u\\left[x.\\diff{\\left[\\log_{}\\left(x+\\frac{1}{x}\\right)\\right]}{x}+\\log_{}\\left(x+\\frac{1}{x}\\right)\\diff{x}{x}\\right]}\\\\\n{\\frac{du}{dx}}&={u\\left[x\\times\\frac{1}{\\left(x+\\frac{1}{x}\\right)}\\times\\left(1-\\frac{1}{x^{2}}\\right)+\\log_{}\\left(x+\\frac{1}{x}\\right)\\right]}\\\\\n{\\frac{du}{dx}}&={u\\left[x\\times\\frac{1}{\\frac{x^{2+1}}{x}}\\times\\left(\\frac{x^{2}-1}{x^{2}}\\right)+\\log_{}\\left(x+\\frac{1}{x}\\right)\\right]}\t\n\\end{align*}","ts":1598269471701,"cs":"Ip0gGLhrW+eNufAeIr51UA==","size":{"width":381,"height":240}}

{"id":"20","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\diff{u}{x}}&={\\left(x+\\frac{1}{x}\\right)^{x}\\left[\\left(\\frac{x^{2}}{x^{2}+1}\\right)\\left(\\frac{x^{2}-1}{x^{2}}\\right)+\\log_{}\\left(x+\\frac{1}{x}\\right)\\right]}\\\\\n{\\diff{u}{x}}&={\\left(x+\\frac{1}{x}\\right)^{x}\\left[\\left(\\frac{x^{2}-1}{x^{2}+1}\\right)+\\log_{}\\left(x+\\frac{1}{x}\\right)\\right]}\t\n\\end{align*}","type":"align*","ts":1598269735527,"cs":"NOsVSCVxlf2UZsUbBzwtDg==","size":{"width":392,"height":82}}

⇒ v = x1+1/x

Taking log both sides:-

⇒ log v = log (x)1+1/x

We know that log xn = n.(log x)

⇒ log v = (1+1/x).log (x)

Differentiate w.r.t. x:-

{"code":"\\begin{align*}\n{\\frac{1}{v}\\diff{v}{x}}&={\\diff{\\left[\\left(1+\\frac{1}{x}\\right).\\log_{}\\left(x\\right)\\right]}{x}}\\\\\n{\\frac{1}{v}\\frac{dv}{dx}}&={\\left(1+\\frac{1}{x}\\right).\\diff{\\left[\\log_{}\\left(x\\right)\\right]}{x}+\\log_{}\\left(x\\right)\\diff{\\left(1+\\frac{1}{x}\\right)}{x}}\\\\\n{\\frac{dv}{dx}}&={v\\left[\\left(1+\\frac{1}{x}\\right).\\frac{1}{x}+\\log_{}\\left(x\\right)\\times\\left(0-\\frac{1}{x^{2}}\\right)\\right]}\\\\\n{\\frac{dv}{dx}}&={v\\left[\\left(\\frac{x+1}{x^{2}}\\right)-\\log_{}\\left(x\\right)\\times\\left(\\frac{1}{x^{2}}\\right)\\right]}\\\\\n{\\frac{dv}{dx}}&={v\\left[\\left(\\frac{1}{x^{2}}\\right)\\times\\left(x+1-\\log_{}x\\right)\\right]}\\\\\n{\\frac{dv}{dx}}&={x^{1+\\frac{1}{x}}\\left(\\frac{x+1-\\log_{}x}{x^{2}}\\right)}\\\\\n{Then\\,\\diff{y}{x}}&={\\diff{u}{x}+\\diff{v}{x}}\\\\\n{\\diff{y}{x}}&={\\left(x+\\frac{1}{x}\\right)^{x}\\left(\\frac{x^{2}-1}{x^{2}+1}+\\log_{}\\left(x+\\frac{1}{x}\\right)\\right)+x^{1+\\frac{1}{x}}\\left(\\frac{x+1-\\log_{}x}{x^{2}}\\right)}\t\n\\end{align*}","type":"align*","id":"19-1","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1598270746025,"cs":"B9CSFCDsPN5o1quw7BosCA==","size":{"width":513,"height":336}}

7. (log x)x + xlog x

Solun:- Let y = (log x)x + xlog x

Differentiate w.r.t. x:-

⇒ y = u + v

⇒ u = (log x)x

Taking log both sides:-

⇒ log u = log [(log x)x]

We know that log xn = n.(log x)

⇒ log u = x.[log (log x)]

Differentiate w.r.t. x:-

{"font":{"size":12,"color":"#000000","family":"Arial"},"id":"21-0","code":"$\\frac{1}{u}\\diff{u}{x}=\\diff{\\left(x.\\left[\\log_{}\\left(\\log_{}x\\right)\\right]\\right)}{x}$","type":"$","ts":1598343413231,"cs":"hlzBKqC9r76JUUNBgoleyA==","size":{"width":177,"height":28}}

We know that:-

{"id":"22-0-0","code":"$\\diff{\\left(u.v\\right)}{x}=u\\diff{v}{x}+v\\diff{u}{x}$","font":{"family":"Arial","color":"#000000","size":12},"type":"$","ts":1598343499540,"cs":"Gyb7xa3pPc/JYUyDqEkVSw==","size":{"width":170,"height":28}}

{"id":"23-0-0","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\frac{1}{u}\\diff{u}{x}}&={x\\diff{\\left[\\log_{}\\left(\\log_{}x\\right)\\right]}{x}+\\log_{}\\left(\\log_{}x\\right)\\diff{x}{x}}\\\\\n{\\diff{u}{x}}&={u\\left[x\\diff{\\left[\\log_{}\\left(\\log_{}x\\right)\\right]}{x}+\\log_{}\\left(\\log_{}x\\right)\\diff{x}{x}\\right]}\\\\\n{\\diff{u}{x}}&={u\\left[x\\times\\frac{1}{\\log_{}x}\\times\\frac{1}{x}+\\log_{}\\left(\\log_{}x\\right)\\right]}\\\\\n{\\diff{u}{x}}&={\\left(\\log_{}x\\right)^{x}\\left[\\frac{1}{\\log_{}x}+\\log_{}\\left(\\log_{}x\\right)\\right]}\\\\\n{\\diff{u}{x}}&={\\left(\\log_{}x\\right)^{x}\\left[\\frac{1+\\log_{}x\\times\\log_{}\\left(\\log_{}x\\right)}{\\log_{}x}\\right]}\\\\\n{\\diff{u}{x}}&={\\left(\\log_{}x\\right)^{x-1}\\left(1+\\log_{}x\\times\\log_{}\\left(\\log_{}x\\right)\\right)}\t\n\\end{align*}","ts":1598344220222,"cs":"9v4x9FJgrjSr0xj/sZsLOA==","size":{"width":297,"height":242}}

⇒ v = xlog x

Taking log both sides:-

⇒ log v = log (x)log x

We know that log xn = n.(log x)

⇒ log v = log x.log x

⇒ log v = (log x)2

Differentiate w.r.t. x:-

{"type":"align*","code":"\\begin{align*}\n{\\frac{1}{v}\\diff{v}{x}}&={\\diff{\\left[\\left(\\log_{}x\\right)^{2}\\right]}{x}}\\\\\n{\\diff{v}{x}}&={v\\left[2\\log_{}x\\times\\frac{1}{x}\\right]}\\\\\n{\\diff{v}{x}}&={x^{\\log_{}x}\\left(\\frac{2\\log_{}x}{x}\\right)}\\\\\n{\\diff{v}{x}}&={2x^{\\log_{}x-1}.\\log_{}x}\\\\\n{\\Rightarrow\\diff{y}{x}}&={\\diff{u}{x}+\\diff{v}{x}}\\\\\n{\\diff{y}{x}}&={\\left(\\log_{}x\\right)^{x-1}\\left(1+\\log_{}x\\times \\log_{}\\left(\\log_{}x\\right)\\right)+2x^{\\log_{}x-1}.\\log_{}x}\t\n\\end{align*}","id":"24","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1598346797833,"cs":"YHhGv6UowguARbVbE5YaVA==","size":{"width":404,"height":244}}

8. (sin x)x + sin-1 √x

Solun:- Let y = (sin x)x + sin-1 √x

Differentiate w.r.t. x:-

⇒ y = u + sin-1 √x

⇒ u = (sin x)x

Taking log both sides:-

⇒ log u = [log (sin x)x]

We know that log xn = n.(log x)

⇒ log u = x.[log (sin x)]

Differentiate w.r.t. x:-

{"font":{"family":"Arial","color":"#000000","size":12},"code":"$\\frac{1}{u}\\diff{u}{x}=\\diff{\\left(x.\\left[\\log_{}\\left(\\sin x\\right)\\right]\\right)}{x}$","type":"$","id":"21-1","ts":1598347216825,"cs":"mJOBkKlSizE5ypHkFOez8Q==","size":{"width":176,"height":28}}

We know that:-

{"id":"22-1","code":"$\\diff{\\left(u.v\\right)}{x}=u\\diff{v}{x}+v\\diff{u}{x}$","font":{"family":"Arial","color":"#000000","size":12},"type":"$","ts":1598343499540,"cs":"HUrfE2pvvk8Hvw9+tF3L2g==","size":{"width":170,"height":28}}

{"id":"25","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\frac{1}{u}\\diff{u}{x}}&={x\\diff{\\left[\\left(\\log_{}\\left(\\sin x\\right)\\right)\\right]}{x}+\\log_{}\\left(\\sin x\\right)\\diff{x}{x}}\\\\\n{\\diff{u}{x}}&={u\\left[x\\diff{\\left[\\left(\\log_{}\\left(\\sin x\\right)\\right)\\right]}{x}+\\log_{}\\left(\\sin x\\right)\\diff{x}{x}\\right]}\\\\\n{\\diff{u}{x}}&={u\\left[x\\times\\frac{1}{\\sin x}\\times \\cos x+\\log_{}\\left(\\sin x\\right)\\right]}\\\\\n{\\diff{u}{x}}&={\\left(\\sin x\\right)^{x}\\left[x\\cot x+\\log_{}\\left(\\sin x\\right)\\right]}\t\n\\end{align*}","type":"align*","ts":1598348229166,"cs":"UapNieALQNVzKcToGd0pRQ==","size":{"width":306,"height":156}}

Then y = u + sin-1 √x

{"id":"26","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{u}{x}+\\diff{\\left(\\sin^{-1}{\\sqrt[]{x}}\\right)}{x}}\\\\\n{\\diff{y}{x}}&={\\left(\\sin x\\right)^{x}\\left[x\\cot x+\\log_{}\\left(\\sin x\\right)\\right]+\\frac{1}{{\\sqrt[]{1-\\left({\\sqrt[]{x}}\\right)^{2}}}}\\times\\frac{1}{2{\\sqrt[]{x}}}}\\\\\n{\\diff{y}{x}}&={\\left(\\sin x\\right)^{x}\\left[x\\cot x+\\log_{}\\left(\\sin x\\right)\\right]+\\frac{1}{2{\\sqrt[]{x-x^{2}}}}}\t\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#222222"},"ts":1598348643177,"cs":"Mdu2+7VyjvVXLbr/uBlreQ==","size":{"width":404,"height":134}}

9. xsin x + (sin x)cos x

Solun:- Let y = xsin x + (sin x)cos x

Differentiate w.r.t. x:-

⇒ y = u + v

⇒ u = xsin x

Taking log both sides:-

⇒ log u = log [xsin x]

We know that log xn = n.(log x)

⇒ log u = sin x.log x

Differentiate w.r.t. x:-

{"font":{"size":12,"color":"#000000","family":"Arial"},"code":"$\\frac{1}{u}\\diff{u}{x}=\\diff{\\left(\\sin x\\times \\log_{}x\\right)}{x}$","id":"21-2-0","type":"$","ts":1598349224323,"cs":"lnGGlmPkqdRjX8miS0Vc/g==","size":{"width":166,"height":28}}

We know that:-

{"id":"22-2-0","code":"$\\diff{\\left(u.v\\right)}{x}=u\\diff{v}{x}+v\\diff{u}{x}$","font":{"family":"Arial","color":"#000000","size":12},"type":"$","ts":1598343499540,"cs":"rs5lL4P8Mzu/uPGIjhN3Kw==","size":{"width":170,"height":28}}

{"type":"$","id":"27-0","font":{"size":12,"color":"#000000","family":"Arial"},"code":"$\\frac{1}{u}\\diff{u}{x}=\\sin x\\diff{\\left(\\log_{}x\\right)}{x}+\\log_{}x\\diff{\\left(\\sin x\\right)}{x}$","ts":1598349357672,"cs":"zHEnUI3ToBQFAfFXDFNWQA==","size":{"width":290,"height":28}}

{"code":"\\begin{align*}\n{\\diff{u}{x}}&={u\\left[\\sin x.\\frac{1}{x}+\\log_{}x.\\cos x\\right]}\\\\\n{\\diff{u}{x}}&={x^{\\sin x}\\left[\\frac{\\sin x}{x}+\\log_{}x.\\cos x\\right]}\t\n\\end{align*}","type":"align*","id":"28-0","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1598349548329,"cs":"V6GslpS3KOzutLEdOsbMHA==","size":{"width":224,"height":80}}

⇒ v = (sin x)cos x

Taking log both sides:-

⇒ log v = log (sin x)cos x

We know that log xn = n.(log x)

⇒ log v = cos x.log (sin x)

Differentiate w.r.t. x:-

{"id":"29","type":"$","code":"$\\frac{1}{v}\\diff{v}{x}=\\diff{\\left[\\cos x\\times \\log_{}\\left(\\sin x\\right)\\right]}{x}$","font":{"family":"Arial","color":"#000000","size":12},"ts":1598349767005,"cs":"p70tDahnRkhn0E6ygYI/rw==","size":{"width":192,"height":28}}

{"type":"align*","id":"30","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\diff{v}{x}}&={v\\left[\\cos x\\times\\diff{\\left[\\log_{}\\left(\\sin x\\right)\\right]}{x}+\\log_{}\\left(\\sin x\\right)\\times\\diff{\\left(\\cos x\\right)}{x}\\right]}\\\\\n{\\diff{v}{x}}&={\\left(\\sin x\\right)^{\\cos x}\\left[\\cos x.\\frac{1}{\\sin x}.\\cos x-\\sin x.\\log_{}\\left(\\sin x\\right)\\right]}\\\\\n{\\diff{v}{x}}&={\\left(\\sin x\\right)^{\\cos x}\\left[\\cos x.\\cot x-\\sin x.\\log_{}\\left(\\sin x\\right)\\right]}\t\n\\end{align*}","ts":1598350041091,"cs":"C1hhhVc7qUCa4rg0X7FW7w==","size":{"width":376,"height":117}}

⇒ y = u + v  Then

{"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{u}{x}+\\diff{v}{x}}\\\\\n{\\diff{y}{x}}&={x^{\\sin x}\\left[\\frac{\\sin x}{x}+\\cos x.\\log_{}x\\right]+\\left(\\sin x\\right)^{\\cos x}\\left[\\cos x.\\cot x-\\sin x.\\log_{}\\left(\\sin x\\right)\\right]}\t\n\\end{align*}","type":"align*","font":{"size":10,"color":"#222222","family":"Arial"},"id":"31","ts":1598350240506,"cs":"9p/Z3jcySgTvPeQ/zAL5YA==","size":{"width":514,"height":74}}

{"code":"$10.\\,x^{x\\cos x}+\\frac{x^{2}+1}{x^{2}-1}$","font":{"size":12,"color":"#434343","family":"Arial"},"id":"32-0","type":"$","ts":1598350688138,"cs":"X2ku/S9xrj2BOKeeJO35ZQ==","size":{"width":148,"height":28}}

Solun:- Let 

{"id":"32-1","font":{"size":12,"family":"Arial","color":"#434343"},"type":"$","code":"$y=x^{x\\cos x}+\\frac{x^{2}+1}{x^{2}-1}$","ts":1598350794363,"cs":"DmazoToJX3w92KfXF/suZQ==","size":{"width":156,"height":28}}

Differentiate w.r.t. x:-

⇒ y = u + v

⇒ u = xxcos x

Taking log both sides:-

⇒ log u = log [xxcos x]

We know that log xn = n.(log x)

⇒ log u = x.cos x.log x

Differentiate w.r.t. x:-

{"font":{"size":12,"family":"Arial","color":"#000000"},"code":"$\\frac{1}{u}\\diff{u}{x}=\\diff{\\left(x.\\cos x.\\log_{}x\\right)}{x}$","type":"$","id":"21-3","ts":1598351002823,"cs":"/DzxJbK5PcJNGk/EBDDe9w==","size":{"width":173,"height":28}}

We know that:-

{"type":"$","code":"$\\diff{\\left(u.v.w\\right)}{x}=u.v\\diff{w}{x}+v.w\\diff{u}{x}+u.w\\diff{v}{x}$","font":{"color":"#000000","family":"Arial","size":12},"id":"22-2-1","ts":1598351073469,"cs":"8+y/PyVxx0tVNa6aEop+iA==","size":{"width":316,"height":28}}

{"id":"33","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\frac{1}{u}\\diff{u}{x}}&={x\\cos x\\diff{\\left(\\log_{}x\\right)}{x}+\\cos x.\\log_{}x\\diff{x}{x}+x.\\log_{}x\\diff{\\left(\\cos x\\right)}{x}}\\\\\n{\\diff{u}{x}}&={u\\left[x\\cos x.\\frac{1}{x}+\\cos x.\\log_{}x-x.\\sin x.\\log_{}x\\right]}\\\\\n{\\diff{u}{x}}&={x^{x\\cos x}\\left[\\cos x\\left(1+\\log_{}x\\right)-x.\\sin x.\\log_{}x\\right]}\t\n\\end{align*}","ts":1598351449379,"cs":"zCuzO+O5iYzTBQMpr3gZRg==","size":{"width":412,"height":113}}

{"type":"$","id":"34","code":"$\\Rightarrow\\,v=\\frac{x^{2}+1}{x^{2}-1}$","font":{"size":12,"color":"#222222","family":"Arial"},"ts":1598351603597,"cs":"h6e0xZR1yS5VAZ9xWF2lLA==","size":{"width":108,"height":28}}

We know that 

{"code":"$\\diff{\\left(\\frac{u}{v}\\right)}{x}=\\frac{vu^{\\prime }-uv^{\\prime }}{v^{2}}$","font":{"family":"Arial","color":"#000000","size":12},"type":"$","id":"35","ts":1598351735679,"cs":"uaB/Cc9ImRHfV07Mfv2RQA==","size":{"width":126,"height":32}}

Differentiate w.r.t. x:-

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"36","code":"\\begin{align*}\n{\\diff{v}{x}}&={\\frac{\\left(x^{2}-1\\right)\\diff{\\left(x^{2}+1\\right)}{x}-\\left(x^{2}+1\\right)\\diff{\\left(x^{2}-1\\right)}{x}}{\\left(x^{2}-1\\right)^{2}}}\\\\\n{\\diff{v}{x}}&={\\frac{2x.\\left(x^{2}-1\\right)-2x.\\left(x^{2}+1\\right)}{\\left(x^{2}-1\\right)^{2}}}\\\\\n{\\diff{v}{x}}&={\\frac{2x.\\left[\\left(x^{2}-1\\right)-\\left(x^{2}+1\\right)\\right]}{\\left(x^{2}-1\\right)^{2}}}\\\\\n{\\diff{v}{x}}&={\\frac{-4x}{\\left(x^{2}-1\\right)^{2}}}\t\n\\end{align*}","type":"align*","ts":1598535014684,"cs":"lDM3U1z+x25ea7FT1KfgHA==","size":{"width":284,"height":193}}

⇒ y = u + v  Then

{"font":{"color":"#222222","size":10,"family":"Arial"},"type":"align*","id":"37","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{u}{x}+\\diff{v}{x}}\\\\\n{\\diff{y}{x}}&={x^{x\\cos x}\\left[\\cos x\\left(1+\\log_{}x\\right)-x\\sin x\\log_{}x\\right]-\\frac{4x}{\\left(x^{2}-1\\right)^{2}}}\t\n\\end{align*}","ts":1598352429594,"cs":"/Tc1iRFUMspQxh6rj99tYQ==","size":{"width":386,"height":77}}

11. (xcos x)x + (xsin x)1/x

Solun:- Let y = (xcos x)x + (xsin x)1/x

Differentiate w.r.t. x:-

⇒ y = u + v

⇒ u = (xcos x)x

Taking log both sides:-

⇒ log u = log (xcos x)x

We know that log xn = n.(log x)

⇒ log u = x.log (xcos x)

Differentiate w.r.t. x:-

{"font":{"size":12,"family":"Arial","color":"#000000"},"id":"21-2-1","type":"$","code":"$\\frac{1}{u}\\diff{u}{x}=\\diff{\\left(x\\times \\log_{}\\left(x\\cos x\\right)\\right)}{x}$","ts":1598353073158,"cs":"7tae1egttLStKjIof0JRgg==","size":{"width":188,"height":28}}

We know that:-

{"id":"22-2-0","code":"$\\diff{\\left(u.v\\right)}{x}=u\\diff{v}{x}+v\\diff{u}{x}$","font":{"family":"Arial","color":"#000000","size":12},"type":"$","ts":1598343499540,"cs":"rs5lL4P8Mzu/uPGIjhN3Kw==","size":{"width":170,"height":28}}

{"font":{"size":12,"color":"#000000","family":"Arial"},"code":"$\\frac{1}{u}\\diff{u}{x}=x\\diff{\\left(\\log_{}\\left(x\\cos x\\right)\\right)}{x}+\\log_{}\\left(x\\cos x\\right)\\diff{\\left(x\\right)}{x}$","id":"27-1","type":"$","ts":1598353877692,"cs":"fKW1lAcRTqqcrLPMjb+IGQ==","size":{"width":341,"height":28}}

{"code":"\\begin{align*}\n{\\diff{u}{x}}&={u\\left[x\\times\\frac{1}{x\\cos x}\\left[x\\left(-\\sin x\\right)+\\cos x\\right]+\\log_{}\\left(x.\\cos x\\right)\\right]}\\\\\n{\\diff{u}{x}}&={\\left(x\\cos x\\right)^{x}\\left[1-x\\tan x+\\log_{}\\left(x.\\cos x\\right)\\right]}\t\n\\end{align*}","id":"28-1","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1598353707907,"cs":"lPwy91ccJYS5PO8kZPBAvw==","size":{"width":381,"height":74}}

⇒ v = (xsin x)1/x

Taking log both sides:-

⇒ log v = log (xsin x)1/x

We know that log xn = n.(log x)

⇒ log v = (1/x).log (x.sin x)

Differentiate w.r.t. x:-

{"code":"\\begin{align*}\n{\\frac{1}{v}\\diff{v}{x}}&={\\frac{1}{x}\\diff{\\left[\\log_{}\\left(x\\sin x\\right)\\right]}{x}+\\log_{}\\left(x\\sin x\\right)\\diff{\\left(\\frac{1}{x}\\right)}{x}}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"id":"38","ts":1598354234031,"cs":"qPBdTmkDg6Mul6L+Ax3CxQ==","size":{"width":316,"height":37}}

{"id":"39","type":"align*","code":"\\begin{align*}\n{\\diff{v}{x}}&={v\\left[\\frac{1}{x}\\times\\frac{1}{x\\sin x}\\times\\left(x\\cos x+\\sin x\\right)-\\frac{1}{x^{2}}\\times \\log_{}\\left(x\\sin x\\right)\\right]}\\\\\n{\\diff{v}{x}}&={\\left(x\\sin x\\right)^{\\frac{1}{x}}\\left[\\frac{\\cot x}{x}+\\frac{1}{x^{2}}-\\frac{\\log_{}\\left(x\\sin x\\right)}{x^{2}}\\right]}\\\\\n{\\diff{v}{x}}&={\\left(x\\sin x\\right)^{\\frac{1}{x}}\\left[\\frac{x\\cot x+1-\\log_{}\\left(x\\sin x\\right)}{x^{2}}\\right]}\t\n\\end{align*}","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1598354820969,"cs":"wtkzeSOgVkfjitx6AFzS6Q==","size":{"width":420,"height":124}}

⇒ y = u + v  Then

{"id":"40","font":{"size":9,"color":"#222222","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{u}{x}+\\diff{v}{x}}\\\\\n{\\diff{y}{x}}&={\\left(x\\cos x\\right)^{x}\\left[1-x\\tan x+\\log_{}\\left(x.\\cos x\\right)\\right]+\\left(x\\sin x\\right)^{\\frac{1}{x}}\\left[\\frac{x\\cot x+1-\\log_{}\\left(x\\sin x\\right)}{x^{2}}\\right]}\t\n\\end{align*}","ts":1598355062020,"cs":"S4L07WEUF+2Kycc7CJuBKw==","size":{"width":569,"height":76}}

Find dy/dx of the functions given in Exercises 12 to 15.

12. xy + yx = 1

Solun:- Let xy + yx = 1

⇒ u + v = 1

⇒ u = xy

Taking log both sides:-

⇒ log u = log xy

We know that log xn = n.(log x)

⇒ log u = y.log x

Differentiate w.r.t. x:-

{"code":"$\\frac{1}{u}\\diff{u}{x}=\\diff{\\left(y.\\left(\\log_{}x\\right)\\right)}{x}$","id":"21-0","font":{"size":12,"color":"#000000","family":"Arial"},"type":"$","ts":1598355482540,"cs":"98Csl6+NlzCv2wH75+wwFQ==","size":{"width":149,"height":28}}

We know that:-

{"id":"22-0-1-0","code":"$\\diff{\\left(u.v\\right)}{x}=u\\diff{v}{x}+v\\diff{u}{x}$","font":{"family":"Arial","color":"#000000","size":12},"type":"$","ts":1598343499540,"cs":"bGbO0KC9NoTvZ2A4hFkPTA==","size":{"width":170,"height":28}}

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","id":"23-0-1-0","code":"\\begin{align*}\n{\\frac{1}{u}\\diff{u}{x}}&={y\\diff{\\left(\\log_{}x\\right)}{x}+\\log_{}x\\diff{y}{x}}\\\\\n{\\diff{u}{x}}&={u\\left[\\frac{y}{x}+\\log_{}x\\diff{y}{x}\\right]}\\\\\n{\\diff{u}{x}}&={x^{y}\\left[\\frac{y}{x}+\\log_{}x\\diff{y}{x}\\right]}\\\\\n{\\diff{u}{x}}&={x^{y-1}\\left[y+x.\\log_{}x\\diff{y}{x}\\right]}\t\n\\end{align*}","ts":1598355783868,"cs":"7OtmNQs9r793rS2+S6kKFA==","size":{"width":208,"height":161}}

⇒ v = yx

Taking log both sides:-

⇒ log v = log yx

We know that log xn = n.(log x)

⇒ log v = x.log y

Differentiate w.r.t. x:-

{"font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\frac{1}{v}\\diff{v}{x}}&={x\\diff{\\left(\\log_{}y\\right)}{x}+\\log_{}y\\diff{x}{x}}\\\\\n{\\diff{v}{x}}&={v\\left[\\frac{x}{y}\\diff{y}{x}+\\log_{}y\\right]}\\\\\n{\\diff{v}{x}}&={y^{x}\\left[\\frac{x}{y}\\diff{y}{x}+\\log_{}y\\right]}\t\n\\end{align*}","type":"align*","id":"41-0","ts":1598357549007,"cs":"DHJE7y9pWNqnkh3eVpmc1A==","size":{"width":204,"height":120}}

Given eq. is u+v = 1

Differentiate w.r.t. x:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\diff{u}{x}+\\diff{v}{x}}&={\\diff{\\left(1\\right)}{x}}\t\n\\end{align*}","id":"42-0","type":"align*","ts":1598358485994,"cs":"tM4RuXzU6mxrlDiLxbdxxA==","size":{"width":125,"height":33}}

{"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{x^{y-1}\\left[y+x.\\log_{}x.\\diff{y}{x}\\right]+y^{x}\\left[\\frac{x}{y}\\diff{y}{x}+\\log_{}y\\right]}&={0}\\\\\n{yx^{y-1}+x^{y}.\\log_{}x.\\diff{y}{x}+x.y^{x-1}\\diff{y}{x}+y^{x}.\\log_{}y}&={0}\t\n\\end{align*}","id":"43-0-0","type":"align*","ts":1598357977315,"cs":"iKHJkjdMZIEJnTy6NZWvPw==","size":{"width":334,"height":74}}

{"font":{"size":12,"color":"#000000","family":"Arial"},"code":"$\\diff{y}{x}=\\frac{-\\left(yx^{y-1}+y^{x}.\\log_{}y\\right)}{\\left(x^{y}.\\log_{}x+x.y^{x-1}\\right)}$","type":"$","id":"44-0","ts":1598358426440,"cs":"3cs9zI0nPWSsG9oexDEgzQ==","size":{"width":176,"height":34}}

13. yx = xy

Solun:- Let yx = xy

⇒ u = v

⇒ u = yx

Taking log both sides:-

⇒ log u = log yx

We know that log xn = n.(log x)

⇒ log u = x.log y

Differentiate w.r.t. x:-

{"font":{"color":"#000000","family":"Arial","size":12},"type":"$","code":"$\\frac{1}{u}\\diff{u}{x}=\\diff{\\left(x.\\left(\\log_{}y\\right)\\right)}{x}$","id":"21-0","ts":1598358812035,"cs":"9rPaZucUA3NRNNQP26Zj3g==","size":{"width":149,"height":28}}

We know that:-

{"id":"22-0-1-1","code":"$\\diff{\\left(u.v\\right)}{x}=u\\diff{v}{x}+v\\diff{u}{x}$","font":{"family":"Arial","color":"#000000","size":12},"type":"$","ts":1598343499540,"cs":"jjaNRc0BHd2mvdEeSMwFVw==","size":{"width":170,"height":28}}

{"code":"\\begin{align*}\n{\\frac{1}{u}\\diff{u}{x}}&={x\\diff{\\left(\\log_{}y\\right)}{x}+\\log_{}y\\diff{x}{x}}\\\\\n{\\diff{u}{x}}&={u\\left[\\frac{x}{y}\\diff{y}{x}+\\log_{}y\\right]}\\\\\n{\\diff{u}{x}}&={y^{x}\\left[\\frac{x}{y}\\diff{y}{x}+\\log_{}y\\right]}\\\\\n{\\diff{u}{x}}&={y^{x-1}\\left[x\\diff{y}{x}+y.\\log_{}y\\right]}\t\n\\end{align*}","id":"23-0-1-1-0","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1598535835124,"cs":"NrLk2QOYAdQAVjL2XwCXpw==","size":{"width":206,"height":161}}

⇒ v = xy

Taking log both sides:-

⇒ log v = log xy

We know that log xn = n.(log x)

⇒ log v = y.log x

Differentiate w.r.t. x:-

{"type":"align*","code":"\\begin{align*}\n{\\frac{1}{v}\\diff{v}{x}}&={y\\diff{\\left(\\log_{}x\\right)}{x}+\\log_{}x\\diff{y}{x}}\\\\\n{\\diff{v}{x}}&={v\\left[\\frac{y}{x}+\\left(\\log_{}x\\right)\\diff{y}{x}\\right]}\\\\\n{\\diff{v}{x}}&={x^{y}\\left[\\frac{y}{x}+\\left(\\log_{}x\\right)\\diff{y}{x}\\right]}\\\\\n{\\diff{v}{x}}&={x^{y-1}\\left[y+x.\\left(\\log_{}x\\right)\\diff{y}{x}\\right]}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"id":"41-1-0","ts":1598359096520,"cs":"1o+yo67ZcZ4CUr6819N7kw==","size":{"width":220,"height":161}}

Given eq. is u = v

Differentiate w.r.t. x:-

{"type":"align*","code":"\\begin{align*}\n{\\diff{u}{x}}&={\\diff{v}{x}}\t\n\\end{align*}","id":"42-1","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1598359125592,"cs":"QGE99Y15TXDest1rhi200w==","size":{"width":70,"height":32}}

{"id":"66","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{y^{x-1}\\left[x\\diff{y}{x}+y.\\log_{}y\\right]}&={x^{y-1}\\left[y+x.\\log_{}x.\\diff{y}{x}\\right]}\\\\\n{x.y^{x-1}.\\diff{y}{x}-x^{y}.\\log_{}x.\\diff{y}{x}}&={y.x^{y-1}-y^{x}.\\log_{}y}\\\\\n{\\diff{y}{x}\\left(x.y^{x-1}-x^{y}.\\log_{}x\\right)}&={y.x^{y-1}-y^{x}.\\log_{}y}\t\n\\end{align*}","type":"align*","ts":1598536025452,"cs":"LLO3jzt9HjHpS5j5abAx6g==","size":{"width":358,"height":112}}

{"id":"44-1-0-0","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{y.x^{y-1}-y^{x}.\\log_{}y}{x.y^{x-1}-x^{y}.\\log_{}x}}\t\n\\end{align*}","type":"align*","ts":1598536056036,"cs":"Qubb9GHp8p/Y7iDz5ybFOw==","size":{"width":172,"height":40}}

Given xy = yx

{"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{y.\\frac{x^{y}}{x}-y^{x}.\\log_{}y}{x.\\frac{y^{x}}{y}-x^{y}.\\log_{}x}}\\\\\n{\\diff{y}{x}}&={\\frac{y.\\frac{y^{x}}{x}-y^{x}.\\log_{}y}{x.\\frac{y^{x}}{y}-y^{x}.\\log_{}x}}\\\\\n{\\diff{y}{x}}&={\\frac{y^{x}}{y^{x}}\\left(\\frac{y.\\frac{1}{x}-\\log_{}y}{x.\\frac{1}{y}-\\log_{}x}\\right)}\\\\\n{\\diff{y}{x}}&={\\frac{y}{x}\\left(\\frac{y-x.\\log_{}y}{x-y.\\log_{}x}\\right)}\t\n\\end{align*}","id":"44-1-0-1","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1598536387498,"cs":"hXOI4E04LNRxb8t9lTMWpw==","size":{"width":180,"height":198}}

14. (cos x)y = (cos y)x

Solun:- Let (cos x)y = (cos y)x

⇒ u = v

⇒ u = (cos x)y

Taking log both sides:-

⇒ log u = log (cos x)y

We know that log xn = n.(log x)

⇒ log u = y.log (cos x)

Differentiate w.r.t. x:-

{"id":"21-0","type":"$","code":"$\\frac{1}{u}\\diff{u}{x}=\\diff{\\left(y.\\left(\\log_{}\\left(\\cos x\\right)\\right)\\right)}{x}$","font":{"size":12,"family":"Arial","color":"#000000"},"ts":1598433461349,"cs":"dV7GsrLcuKWKY03wBN0jPQ==","size":{"width":180,"height":28}}

We know that:-

{"id":"22-0-1-2","code":"$\\diff{\\left(u.v\\right)}{x}=u\\diff{v}{x}+v\\diff{u}{x}$","font":{"family":"Arial","color":"#000000","size":12},"type":"$","ts":1598343499540,"cs":"TIMUIinyO76o4ZyW/W0O7w==","size":{"width":170,"height":28}}

{"code":"\\begin{align*}\n{\\frac{1}{u}\\diff{u}{x}}&={y\\diff{\\left(\\log_{}\\left(\\cos x\\right)\\right)}{x}+\\log_{}\\left(\\cos x\\right)\\diff{y}{x}}\\\\\n{\\diff{u}{x}}&={u\\left[y\\diff{\\left(\\log_{}\\left(\\cos x\\right)\\right)}{x}+\\log_{}\\left(\\cos x\\right)\\diff{y}{x}\\right]}\\\\\n{\\diff{u}{x}}&={\\left(\\cos x\\right)^{y}\\left[y\\times\\frac{1}{\\cos x}\\times\\left(-\\sin x\\right)+\\log_{}\\left(\\cos x\\right)\\diff{y}{x}\\right]}\\\\\n{\\diff{u}{x}}&={\\left(\\cos x\\right)^{y}\\left[-y.\\tan x+\\log_{}\\left(\\cos x\\right)\\diff{y}{x}\\right]}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"23-0-1-1-1","ts":1598433866427,"cs":"MopkkH9hT1MU2rLWhwrdcw==","size":{"width":384,"height":162}}

⇒ v = (cos y)x

Taking log both sides:-

⇒ log v = (cos y)x

We know that log xn = n.(log x)

⇒ log v = x.log (cos y)

Differentiate w.r.t. x:-

{"code":"\\begin{align*}\n{\\frac{1}{v}\\diff{v}{x}}&={x\\diff{\\left(\\log_{}\\left(\\cos y\\right)\\right)}{x}+\\log_{}\\left(\\cos y\\right)\\diff{x}{x}}\\\\\n{\\diff{v}{x}}&={v\\left[x\\times\\frac{1}{\\cos y}\\times\\left(-\\sin y\\right)\\diff{y}{x}+\\log_{}\\left(\\cos y\\right)\\diff{x}{x}\\right]}\\\\\n{\\diff{v}{x}}&={\\left(\\cos y\\right)^{x}\\left[-x.\\tan y\\diff{y}{x}+\\log_{}\\left(\\cos y\\right)\\right]}\\\\\n{\\diff{v}{x}}&={\\left(\\cos y\\right)^{x}\\left[-x.\\tan y\\diff{y}{x}+\\log_{}\\left(\\cos y\\right)\\right]}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"id":"41-1-1","type":"align*","ts":1598434247868,"cs":"4JyVQh+PfIvFEJFFlA0Skw==","size":{"width":362,"height":161}}

Given eq. is u = v

Differentiate w.r.t. x:-

{"type":"align*","code":"\\begin{align*}\n{\\diff{u}{x}}&={\\diff{v}{x}}\t\n\\end{align*}","id":"42-1","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1598359125592,"cs":"QGE99Y15TXDest1rhi200w==","size":{"width":70,"height":32}}

{"code":"\\begin{align*}\n{\\left(\\cos x\\right)^{y}\\left[-y\\tan x+\\log_{}\\left(\\cos x\\right)\\diff{y}{x}\\right]}&={\\left(\\cos y\\right)^{x}\\left[-x\\tan y\\diff{y}{x}+\\log_{}\\left(\\cos y\\right)\\right]}\t\n\\end{align*}","type":"align*","id":"43-0-1-1","font":{"color":"#000000","family":"Arial","size":10},"ts":1598436109109,"cs":"2XxLWl+B2EyziLTyJSYkBA==","size":{"width":490,"height":37}}

{"id":"43-0-1-1","type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}\\left(\\left(\\cos x\\right)^{y}.\\log_{}\\left(\\cos x\\right)+x.\\tan y.\\left(\\cos y\\right)^{x}\\right)}&={\\left(\\cos y\\right)^{x}.\\log_{}\\left(\\cos y\\right)+y\\tan x.\\left(\\cos x\\right)^{y}}\t\n\\end{align*}","font":{"size":8,"color":"#000000","family":"Arial"},"ts":1598436283695,"cs":"ZbVjZfP7L0Y4utAFE3jr3Q==","size":{"width":476,"height":28}}

{"font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{\\left(\\cos y\\right)^{x}.\\log_{}\\left(\\cos y\\right)+y\\tan x.\\left(\\cos x\\right)^{y}}{\\left(\\cos x\\right)^{y}.\\log_{}\\left(\\cos x\\right)+x.\\tan y.\\left(\\cos y\\right)^{x}}}\t\n\\end{align*}","id":"44-1-1","type":"align*","ts":1598436361147,"cs":"RfKdQkh2gCgghjn/7hH+ig==","size":{"width":304,"height":40}}

Given (cos x)y = (cos y)x

{"id":"44-1-1","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{\\left(\\cos y\\right)^{x}.\\log_{}\\left(\\cos y\\right)+y\\tan x.\\left(\\cos y\\right)^{x}}{\\left(\\cos y\\right)^{x}.\\log_{}\\left(\\cos x\\right)+x.\\tan y.\\left(\\cos y\\right)^{x}}}\\\\\n{\\diff{y}{x}}&={\\frac{\\left(\\cos y\\right)^{x}}{\\left(\\cos y\\right)^{x}}\\left(\\frac{\\log_{}\\left(\\cos y\\right)+y\\tan x}{\\log_{}\\left(\\cos x\\right)+x.\\tan y}\\right)}\\\\\n{\\diff{y}{x}}&={\\left(\\frac{\\log_{}\\left(\\cos y\\right)+y\\tan x}{\\log_{}\\left(\\cos x\\right)+x.\\tan y}\\right)}\t\n\\end{align*}","type":"align*","ts":1598436763513,"cs":"6KkPpn/HLCwAnzS4NcX7Vg==","size":{"width":304,"height":126}}

15. xy = e(x-y)

Solun:- Let xy = e(x-y)

Differentiate w.r.t. x:-

{"code":"$\\diff{\\left(xy\\right)}{x}=\\diff{\\left[e^{\\left(x-y\\right)}\\right]}{x}$","font":{"family":"Arial","size":12,"color":"#000000"},"id":"46","type":"$","ts":1598437128394,"cs":"ePWMwKkFqahiqOp3WDqhCA==","size":{"width":133,"height":32}}

We know that

{"type":"$","font":{"family":"Arial","size":12,"color":"#000000"},"id":"47","code":"$\\diff{\\left(u.v\\right)}{x}=u\\diff{v}{x}+v\\diff{u}{x}$","ts":1598437289964,"cs":"PlHQQzd+sSutyE0no3kAQg==","size":{"width":170,"height":28}}

{"code":"\\begin{align*}\n{x\\diff{y}{x}+y\\diff{x}{x}}&={e^{\\left(x-y\\right)}.\\diff{\\left(x-y\\right)}{x}}\t\n\\end{align*}","id":"48","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","ts":1598437540667,"cs":"IYEU2gvpZU8icUfr+mzppQ==","size":{"width":216,"height":33}}

{"code":"\\begin{align*}\n{x\\diff{y}{x}+y}&={e^{\\left(x-y\\right)}.\\left(1-\\diff{y}{x}\\right)}\t\n\\end{align*}","id":"49","font":{"family":"Arial","size":10,"color":"#000000"},"type":"align*","ts":1598437802188,"cs":"Nl59achOLU+uOkAOqx2XyA==","size":{"width":201,"height":37}}

{"font":{"color":"#000000","size":12,"family":"Arial"},"type":"$","code":"$\\diff{y}{x}=\\frac{e^{\\left(x-y\\right)}-y}{x+e^{\\left(x-y\\right)}}$","id":"50","ts":1598437867644,"cs":"tzmZi+gbyVRPC2ZrsWnGyg==","size":{"width":114,"height":32}}

Given xy = e(x-y)

{"font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{xy-y}{x+xy}}\\\\\n{\\diff{y}{x}}&={\\frac{y\\left(x-1\\right)}{x\\left(y+1\\right)}}\t\n\\end{align*}","type":"align*","id":"51","ts":1598438134679,"cs":"iePHIQeL9kgy73V3ChgNig==","size":{"width":108,"height":78}}

16. Find the derivative of the function given by f(x) = (1+x)(1+x2)(1+x4)(1+x8) and hence find f’(1).

Solun:- Let f(x) = (1+x)(1+x2)(1+x4)(1+x8)

Taking log both side:-

log [f(x)] = log (1+x)+log (1+x2)+log (1+x4)+log (1+x8)

Differentiate w.r.t. x:-

{"font":{"color":null,"size":10,"family":"Arial"},"code":"$\\diff{\\left[\\log_{}f\\left(x\\right)\\right]}{x}=\\diff{\\left[log (1+x)+log (1+x^{2})+log (1+x^{4})+log (1+x^{8})\\right]}{x}$","id":"52","type":"$","ts":1598438986068,"cs":"MLHe9UzUcKfcIKJmT8UfFA==","size":{"width":314,"height":24}}

{"code":"$\\frac{1}{f\\left(x\\right)}.f^{\\prime }\\left(x\\right)=\\frac{1}{1+x}+\\frac{2x}{1+x^{2}}+\\frac{4x^{3}}{1+x^{4}}+\\frac{8x^{7}}{1+x^{8}}$","type":"$","id":"53","font":{"family":"Arial","size":12,"color":"#000000"},"ts":1598439469249,"cs":"VzZzdqWzyy+sqXBZOlYWIA==","size":{"width":352,"height":30}}

{"code":"$f^{\\prime }\\left(x\\right)=\\left(1+x\\right)\\left(1+x^{2}\\right)\\left(1+x^{4}\\right)\\left(1+x^{8}\\right)\\left[\\frac{1}{1+x}+\\frac{2x}{1+x^{2}}+\\frac{4x^{3}}{1+x^{4}}+\\frac{8x^{7}}{1+x^{8}}\\right]$","font":{"size":10,"family":"Arial","color":"#000000"},"id":"54","type":"$","ts":1598439656193,"cs":"F91e8sHrqQcKlhJ0U46E1g==","size":{"width":473,"height":28}}

Put x = 1

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"55","type":"align*","code":"\\begin{align*}\n{f^{\\prime }\\left(1\\right)}&={16\\left(\\frac{1}{2}+\\frac{2}{2}+\\frac{4}{2}+\\frac{8}{2}\\right)}\\\\\n{f^{\\prime }\\left(1\\right)}&={16\\left(\\frac{1}{2}+1+2+4\\right)}\\\\\n{f^{\\prime }\\left(1\\right)}&={120}\t\n\\end{align*}","ts":1598440190622,"cs":"Np2ZrSXqeGm+s9Qlc7p+JA==","size":{"width":212,"height":101}}

17. Differentiate (x2-5x+8). (x3+7x+9) in three ways mentioned below:

(i) by using product rule

Solun:- Let y  = (x2-5x+8). (x3+7x+9)

Differentiate w.r.t. x:-

{"font":{"size":12,"color":"#000000","family":"Arial"},"code":"$\\diff{y}{x}=\\diff{\\left[\\left(x^{2}-5x+8\\right)\\left(x^{3}+7x+9\\right)\\right]}{x}$","id":"56","type":"$","ts":1598441337571,"cs":"lNC0O/Q8IIL5HskCHhGCkQ==","size":{"width":225,"height":30}}

We know that

{"code":"$\\diff{\\left(u.v\\right)}{x}=u\\diff{v}{x}+v\\diff{u}{x}$","id":"57","type":"$","font":{"family":"Arial","color":"#000000","size":12},"ts":1598441387598,"cs":"O+w7LgPQA5URlIEpgCzDMA==","size":{"width":170,"height":28}}

{"id":"58","font":{"family":"Arial","size":8,"color":"#000000"},"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\left(x^{2}-5x+8\\right)\\diff{\\left(x^{3}+7x+9\\right)}{x}+\\left(x^{3}+7x+9\\right)\\diff{\\left(x^{2}-5x+8\\right)}{x}}\\\\\n{\\diff{y}{x}}&={\\left(x^{2}-5x+8\\right)\\left(3x^{2}+7\\right)+\\left(x^{3}+7x+9\\right)\\left(2x-5\\right)}\\\\\n{\\diff{y}{x}}&={3x^{4}+7x^{2}-15x^{3}-35x+24x^{2}+56+2x^{4}-5x^{3}+14x^{2}-35x+18x-45}\\\\\n{\\diff{y}{x}}&={5x^{4}-20x^{3}+45x^{2}-52x+11}\t\n\\end{align*}","type":"align*","ts":1598442183793,"cs":"tObGbzp/1w8/RqnBNkhNKA==","size":{"width":473,"height":128}}

(ii) by expanding the product to obtain a single polynomial

Solun:- Let y  = (x2-5x+8). (x3+7x+9)

⇒ y  = x5+7x3+9x2-5x4-35x2-45x+8x3+56x+72

Differentiate w.r.t. x:-

{"type":"align*","id":"59-0","font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{\\left[x^{5}-5x^{4}+15x^{3}-26x^{2}-11x+72\\right]}{x}}\\\\\n{\\diff{y}{x}}&={5x^{4}-20x^{3}+45x^{2}-52x+11}\t\n\\end{align*}","ts":1598537009060,"cs":"4I7JDcjfZhMUobKC12b9bg==","size":{"width":313,"height":74}}

(iii) by logarithmic differentiation

Solun:- Let y  = (x2-5x+8). (x3+7x+9)

Taking log both side

⇒ log y = log [(x2-5x+8). (x3+7x+9)]

⇒ log y = log (x2-5x+8) + log (x3+7x+9)

Differentiate w.r.t. x:-

{"id":"59-1-0","code":"\\begin{align*}\n{\\frac{1}{y}\\diff{y}{x}}&={\\diff{\\left[log (x^{2}-5x+8) + log (x^{3}+7x+9)\\right]}{x}}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\frac{1}{x^{2}-5x+8}\\times\\left(2x-5\\right)+\\frac{1}{x^{3}+7x+9}\\times\\left(3x^{2}+7\\right)}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\frac{2x-5}{x^{2}-5x+8}+\\frac{3x^{2}+7}{x^{3}+7x+9}}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\frac{\\left(2x-5\\right)\\left(x^{3}+7x+9\\right)+\\left(3x^{2}+7\\right)\\left(x^{2}-5x+8\\right)}{\\left(x^{2}-5x+8\\right).\\left(x^{3}+7x+9\\right)}}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","ts":1598537086224,"cs":"sJxnBUI9Sd5RLr8mTxWnag==","size":{"width":412,"height":169}}

{"code":"\\begin{align*}\n{\\diff{y}{x}}&={y.\\left[\\frac{\\left(2x-5\\right)\\left(x^{3}+7x+9\\right)+\\left(3x^{2}+7\\right)\\left(x^{2}-5x+8\\right)}{\\left(x^{2}-5x+8\\right).\\left(x^{3}+7x+9\\right)}\\right]}\\\\\n{\\diff{y}{x}}&={\\left(x^{2}-5x+8\\right).\\left(x^{3}+7x+9\\right)\\left[\\frac{\\left(2x-5\\right)\\left(x^{3}+7x+9\\right)+\\left(3x^{2}+7\\right)\\left(x^{2}-5x+8\\right)}{\\left(x^{2}-5x+8\\right).\\left(x^{3}+7x+9\\right)}\\right]}\\\\\n{\\diff{y}{x}}&={\\left[\\left(2x-5\\right)\\left(x^{3}+7x+9\\right)+\\left(3x^{2}+7\\right)\\left(x^{2}-5x+8\\right)\\right]}\\\\\n{\\diff{y}{x}}&={5x^{4}-20x^{3}+45x^{2}-52x-11}\t\n\\end{align*}","id":"59-1-1","font":{"color":"#000000","family":"Arial","size":2.469798657718121},"type":"align*","ts":1598537180390,"cs":"KV0uhDs5oEVE2+1/4bLwsw==","size":{"width":508,"height":46}}

{"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\left(x^{2}-5x+8\\right).\\left(x^{3}+7x+9\\right)\\left[\\frac{\\left(2x-5\\right)\\left(x^{3}+7x+9\\right)+\\left(3x^{2}+7\\right)\\left(x^{2}-5x+8\\right)}{\\left(x^{2}-5x+8\\right).\\left(x^{3}+7x+9\\right)}\\right]}\\\\\n{\\diff{y}{x}}&={\\left[\\left(2x-5\\right)\\left(x^{3}+7x+9\\right)+\\left(3x^{2}+7\\right)\\left(x^{2}-5x+8\\right)\\right]}\\\\\n{\\diff{y}{x}}&={5x^{4}-20x^{3}+45x^{2}-52x+11}\t\n\\end{align*}","font":{"color":"#000000","size":8,"family":"Arial"},"type":"align*","id":"59-1-1","ts":1598537295613,"cs":"QjqBXSztsfBPeKOwsMdriw==","size":{"width":508,"height":104}}

18. If u, v and w are functions of x, then show that

{"id":"60","code":"$\\diff{\\left(u.v.w\\right)}{x}=\\diff{\\left(u\\right)}{x}.v.w+u.\\diff{\\left(v\\right)}{x}.w+u.v.\\diff{\\left(w\\right)}{x}$","type":"$","font":{"family":"Arial","color":"#000000","size":11},"ts":1598446600656,"cs":"pe+M064EXGaA44Nkbk0Gog==","size":{"width":333,"height":24}}

in two ways- first by repeated application of product rule, second by logarithmic differentiation.

Solun:- Let y  = (u.v).w

By Product Rule:-

Differentiate w.r.t. x:-

{"font":{"family":"Arial","size":10,"color":"#000000"},"id":"61","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{\\left[\\left(u.v\\right).w\\right]}{x}}\\\\\n{\\diff{y}{x}}&={\\left(u.v\\right).\\diff{w}{x}+w.\\diff{\\left(u.v\\right)}{x}}\\\\\n{\\diff{y}{x}}&={\\left(u.v\\right).\\diff{w}{x}+w.\\left[u\\diff{v}{x}+v\\diff{u}{x}\\right]}\\\\\n{\\diff{y}{x}}&={\\left(u.v\\right).\\diff{w}{x}+u.w.\\diff{v}{x}+v.w.\\diff{u}{x}}\t\n\\end{align*}","type":"align*","ts":1598446962770,"cs":"0dnmE4pjn7HDgr0cLPE80g==","size":{"width":272,"height":152}}

{"code":"$\\diff{\\left(u.v.w\\right)}{x}=\\left(u.v\\right).\\diff{w}{x}+u.w.\\diff{v}{x}+v.w.\\diff{u}{x}$","id":"62","type":"$","font":{"size":12,"color":"#000000","family":"Arial"},"ts":1598447145656,"cs":"v/16z/emjZEvjJ7Tlx9amA==","size":{"width":358,"height":28}}

Hence R.H.S = L.H.S

By logarithmic differentiation:-

⇒ y  = u.v.w

⇒ log y = log (u.v.w)

⇒ log y = log u + log v + log w

{"font":{"size":10,"color":"#222222","family":"Arial"},"code":"\\begin{align*}\n{\\frac{1}{y}\\diff{y}{x}}&={\\diff{\\left(log u + log v + log w\\right)}{x}}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\frac{1}{u}\\diff{u}{x}+\\frac{1}{v}\\diff{v}{x}+\\frac{1}{w}\\diff{w}{x}}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\frac{v.w.\\diff{u}{x}+u.w.\\diff{v}{x}+u.v.\\diff{w}{x}}{u.v.w}}\t\n\\end{align*}","type":"align*","id":"63","ts":1598447752500,"cs":"aN1yyqyTPFmBBJt/TW+dpA==","size":{"width":264,"height":124}}

{"font":{"family":"Arial","color":"#222222","size":10},"id":"64","type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={y\\left[\\frac{v.w.\\diff{u}{x}+u.w.\\diff{v}{x}+u.v.\\diff{w}{x}}{u.v.w}\\right]}\\\\\n{\\diff{y}{x}}&={\\left(u.v.w\\right)\\left[\\frac{v.w.\\diff{u}{x}+u.w.\\diff{v}{x}+u.v.\\diff{w}{x}}{u.v.w}\\right]}\\\\\n{\\diff{y}{x}}&={v.w.\\diff{u}{x}+u.w.\\diff{v}{x}+u.v.\\diff{w}{x}}\t\n\\end{align*}","ts":1598447788065,"cs":"PpxDhbvMaKrDhM/Q8L3QtA==","size":{"width":322,"height":136}}

{"font":{"family":"Arial","color":"#222222","size":12},"type":"$","id":"65","code":"$\\diff{\\left(u.v.w\\right)}{x}=v.w.\\diff{u}{x}+u.w.\\diff{v}{x}+u.v.\\diff{w}{x}$","ts":1598447693053,"cs":"WU9rqsn5gONb7JovSWh1Tg==","size":{"width":342,"height":28}}

Hence R.H.S = L.H.S

So, in both cases:- R.H.S = L.H.S


SEE ALSO:-

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