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 Miscellaneous Exercise

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

1. (3x2 - 9x + 5)9

Solun:- Let y = (3x2 - 9x + 5)9

Differentiate w.r.t. x:-

{"id":"1-0-0","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{\\left[(3x^{2}- 9x + 5)^{9}\\right]}{x}}\\\\\n{\\diff{y}{x}}&={9\\left(3x^{2}- 9x + 5\\right)^{8}\\times\\left(6x-9\\right)}\\\\\n{\\diff{y}{x}}&={27\\left(3x^{2}- 9x + 5\\right)^{8}\\times\\left(2x-3\\right)}\t\n\\end{align*}","ts":1598959780562,"cs":"/ys0qeVFlvaVKwIbl0KUUg==","size":{"width":245,"height":112}}

2. sin3x + cos6x

Solun:- Let y = sin3x + cos6x

Differentiate w.r.t. x:-

{"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{\\left[\\sin ^{3}x + \\cos ^{6}x\\right]}{x}}\\\\\n{\\diff{y}{x}}&={\\left(3\\sin^{2}x.\\cos x-6\\cos^{5}x.\\sin x\\right)}\\\\\n{\\diff{y}{x}}&={3\\sin x.\\cos x\\left(\\sin x-2\\cos^{4}x\\right)}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"type":"align*","id":"1-0-1-0","ts":1598960335985,"cs":"h7iWeE6q/DEFOYrU8W1m/w==","size":{"width":256,"height":112}}

3. (5x)3cos 2x

Solun:- Let y = (5x)3cos 2x

Taking log both sides

⇒ log y = log (5x)3cos 2x

We know that log (x)n = nlog x

⇒ log y = (3cos 2x).(log 5x)

Differentiate w.r.t. x:-

{"code":"\\begin{align*}\n{\\frac{1}{y}\\diff{y}{x}}&={\\diff{\\left[\\left(3\\cos2x\\right).\\left(\\log_{}5x\\right)\\right]}{x}}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\left(3\\cos2x\\right).\\diff{\\left(\\log_{}5x\\right)}{x}+\\left(\\log_{}5x\\right).\\diff{\\left(3\\cos2x\\right)}{x}}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\left(3\\cos2x\\right).\\frac{5}{5x}+\\left(\\log_{}5x\\right).\\left(-6\\sin2x\\right)}\t\n\\end{align*}","id":"1-0-1-1-0","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1598964126262,"cs":"v32X6vwxrjX0kJPo/CjxLw==","size":{"width":358,"height":120}}

{"type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"id":"2-0","code":"\\begin{align*}\n{\\diff{y}{x}}&={y\\left[\\frac{\\left(3\\cos2x\\right)}{x}-6\\sin2x.\\log_{}5x\\right]}\\\\\n{\\diff{y}{x}}&={\\left(5x\\right)^{3\\cos2x}\\left[\\frac{\\left(3\\cos2x\\right)}{x}-6\\sin2x.\\log_{}5x\\right]}\t\n\\end{align*}","ts":1598964691901,"cs":"B9Udp/Frzc3lbrnB4PWG2g==","size":{"width":316,"height":80}}

4. sin-1(x√x),  0 ≤ x ≤ 1

Solun:- Let y = sin-1(x√x)

Differentiate w.r.t. x:-

{"type":"align*","id":"1-0-1-1-1-0","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{\\left[\\sin^{-1}x{\\sqrt[]{x}}\\right]}{x}}\\\\\n{\\diff{y}{x}}&={\\frac{1}{{\\sqrt[]{1-\\left(x{\\sqrt[]{x}}\\right)^{2}}}}\\left(x\\diff{\\left({\\sqrt[]{x}}\\right)}{x}+{\\sqrt[]{x}}\\diff{x}{x}\\right)}\\\\\n{\\diff{y}{x}}&={\\frac{1}{{\\sqrt[]{1-x^{3}}}}\\left(\\frac{x}{2{\\sqrt[]{x}}}+{\\sqrt[]{x}}\\right)}\\\\\n{\\diff{y}{x}}&={\\frac{1}{{\\sqrt[]{1-x^{3}}}}\\left(\\frac{x+2x}{2{\\sqrt[]{x}}}\\right)}\\\\\n{\\diff{y}{x}}&={\\frac{1}{{\\sqrt[]{1-x^{3}}}}\\left(\\frac{3x}{2{\\sqrt[]{x}}}\\right)}\\\\\n{\\diff{y}{x}}&={\\frac{3}{2}.\\frac{{\\sqrt[]{x}}}{{\\sqrt[]{1-x^{3}}}}}\\\\\n{\\diff{y}{x}}&={\\frac{3}{2}{\\sqrt[]{\\frac{x}{1-x^{3}}}}}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"ts":1598965258710,"cs":"oMOCApPrbgMHIc2GMzqffg==","size":{"width":302,"height":318}}

{"code":"$5.\\,\\frac{\\cos^{-1}\\frac{x}{2}}{{\\sqrt[]{2x+7}}},\\,-2<x<2$","type":"$","font":{"size":12,"family":"Arial","color":null},"id":"3","ts":1598965484670,"cs":"59SobGcaYI+tbN9+madtYQ==","size":{"width":198,"height":36}}

Solun:- Let 

{"font":{"size":12,"color":"#000000","family":"Arial"},"type":"$","code":"$y=\\frac{\\cos^{-1}\\frac{x}{2}}{{\\sqrt[]{2x+7}}}$","id":"4-0","ts":1598965837822,"cs":"DQuCqurP6KExoetw+BgXkg==","size":{"width":93,"height":36}}

Differentiate w.r.t. x:-

{"font":{"color":"#000000","family":"Arial","size":10},"id":"1-0-1-1-1-1-0-0","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{\\left[\\frac{\\cos^{-1}\\frac{x}{2}}{{\\sqrt[]{2x+7}}}\\right]}{x}}\\\\\n{\\diff{y}{x}}&={\\frac{{\\sqrt[]{2x+7}}\\times\\diff{\\left(\\cos^{-1}\\frac{x}{2}\\right)}{x}-\\cos^{-1}\\frac{x}{2}\\diff{\\left({\\sqrt[]{2x+7}}\\right)}{x}}{\\left({\\sqrt[]{2x+7}}\\right)^{2}}}\\\\\n{\\diff{y}{x}}&={\\frac{{\\sqrt[]{2x+7}}\\times\\frac{\\left(-1\\right)}{2{\\sqrt[]{1-\\left(\\frac{x}{2}\\right)^{2}}}}-\\cos^{-1}\\frac{x}{2}\\times\\frac{2}{2{\\sqrt[]{2x+7}}}}{\\left({\\sqrt[]{2x+7}}\\right)^{2}}}\\\\\n{\\diff{y}{x}}&={\\frac{{\\sqrt[]{2x+7}}\\times\\frac{\\left(-2\\right)}{2{\\sqrt[]{4-x^{2}}}}-\\cos^{-1}\\frac{x}{2}\\times\\frac{2}{2{\\sqrt[]{2x+7}}}}{\\left({\\sqrt[]{2x+7}}\\right)^{2}}}\\\\\n{\\diff{y}{x}}&={\\frac{-{\\sqrt[]{2x+7}}\\times{\\sqrt[]{2x+7}}-{\\sqrt[]{4-x^{2}}}\\times\\cos^{-1}\\frac{x}{2}}{\\left(2x+7\\right).{\\sqrt[]{2x+7}}.{\\sqrt[]{4-x^{2}}}}}\t\n\\end{align*}","type":"align*","ts":1599042791486,"cs":"QPkyKBHnC/ReTCBwLklqDw==","size":{"width":346,"height":320}}

{"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{-{\\sqrt[]{2x+7}}\\times{\\sqrt[]{2x+7}}-{\\sqrt[]{4-x^{2}}}\\times\\cos^{-1}\\frac{x}{2}}{\\left(2x+7\\right).{\\sqrt[]{2x+7}}.{\\sqrt[]{4-x^{2}}}}}\\\\\n{\\diff{y}{x}}&={\\frac{-\\left(2x+7\\right)-{\\sqrt[]{4-x^{2}}}\\times\\cos^{-1}\\frac{x}{2}}{\\left(2x+7\\right).{\\sqrt[]{2x+7}}.{\\sqrt[]{4-x^{2}}}}}\\\\\n{\\diff{y}{x}}&={\\frac{-\\left(2x+7\\right)}{\\left(2x+7\\right).{\\sqrt[]{2x+7}}.{\\sqrt[]{4-x^{2}}}}-\\frac{{\\sqrt[]{4-x^{2}}}\\times\\cos^{-1}\\frac{x}{2}}{\\left(2x+7\\right).{\\sqrt[]{2x+7}}.{\\sqrt[]{4-x^{2}}}}}\\\\\n{\\diff{y}{x}}&={-\\left(\\frac{1}{{\\sqrt[]{2x+7}}.{\\sqrt[]{4-x^{2}}}}+\\frac{\\cos^{-1}\\frac{x}{2}}{\\left(2x+7\\right)^{\\frac{3}{2}}}\\right)}\t\n\\end{align*}","id":"5-0","type":"align*","ts":1599042694149,"cs":"l3++5G/SZjfbp1MxquDP2g==","size":{"width":448,"height":202}}

{"code":"$6.\\,\\cot^{-1}\\left[\\frac{{\\sqrt[]{1+\\sin x}}+{\\sqrt[]{1-\\sin x}}}{{\\sqrt[]{1+\\sin x}}-{\\sqrt[]{1-\\sin x}}}\\right],\\,0<x<\\frac{\\Pi}{2}$","font":{"family":"Arial","color":null,"size":12},"type":"$","id":"3","ts":1599045021649,"cs":"wCe8FzQCtI6zgJEFt3k9SQ==","size":{"width":334,"height":36}}

Solun:- Let

{"type":"align*","id":"4-1","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{y}&={\\cot^{-1}\\left[\\frac{{\\sqrt[]{1+\\sin x}}+{\\sqrt[]{1-\\sin x}}}{{\\sqrt[]{1+\\sin x}}-{\\sqrt[]{1-\\sin x}}}\\right]}\\\\\n{y}&={\\cot^{-1}\\left[\\left(\\frac{{\\sqrt[]{1+\\sin x}}+{\\sqrt[]{1-\\sin x}}}{{\\sqrt[]{1+\\sin x}}-{\\sqrt[]{1-\\sin x}}}\\right)\\times\\left(\\frac{{\\sqrt[]{1+\\sin x}}+{\\sqrt[]{1-\\sin x}}}{{\\sqrt[]{1+\\sin x}}+{\\sqrt[]{1-\\sin x}}}\\right)\\right]}\\\\\n{y}&={\\cot^{-1}\\left[\\left(\\frac{\\left({\\sqrt[]{1+\\sin x}}+{\\sqrt[]{1-\\sin x}}\\right)^{2}}{\\left(1+\\sin x\\right)-\\left(1-\\sin x\\right)}\\right)\\right]}\\\\\n{y}&={\\cot^{-1}\\left[\\frac{\\left(1+\\sin x+1-\\sin x+2{\\sqrt[]{1+\\sin x}}.{\\sqrt[]{1-\\sin x}}\\right)}{2\\sin x}\\right]}\\\\\n{y}&={\\cot^{-1}\\left[\\frac{\\left(2+2{\\sqrt[]{1+\\sin x}}.{\\sqrt[]{1-\\sin x}}\\right)}{2\\sin x}\\right]}\\\\\n{y}&={\\cot^{-1}\\left[\\frac{\\left(1+{\\sqrt[]{1-\\sin ^{2}x}}\\right)}{\\sin x}\\right]}\\\\\n{y}&={\\cot^{-1}\\left[\\frac{\\left(1+\\cos x\\right)}{\\sin x}\\right]}\\\\\n{y}&={\\cot^{-1}\\left[\\frac{2\\cos ^{2}\\frac{x}{2}}{2\\sin \\frac{x}{2}.\\cos\\frac{x}{2}}\\right]}\\\\\n{y}&={\\cot^{-1}\\left(\\cot\\frac{x}{2}\\right)}\\\\\n{y}&={\\frac{x}{2}}\t\n\\end{align*}","ts":1599044541585,"cs":"ai+yeWtwCo1Ppk5rT6XKWQ==","size":{"width":492,"height":541}}

Differentiate w.r.t. x:-

{"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{\\left[\\frac{x}{2}\\right]}{x}}\\\\\n{\\diff{y}{x}}&={\\frac{1}{2}}\t\n\\end{align*}","type":"align*","id":"1-0-1-1-1-1-1","font":{"color":"#000000","family":"Arial","size":10},"ts":1599044624646,"cs":"t3d7iuDpVXE0qd3a9YHZCw==","size":{"width":88,"height":74}}

7. (log x)log x ,  x > 1

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

Taking log both sides:-

⇒ log y = log (log x)log x

We know that log (x)n = nlog x

⇒ log y = (log x).(log (log x))

Differentiate w.r.t. x:-

{"code":"\\begin{align*}\n{\\frac{1}{y}\\diff{y}{x}}&={\\diff{\\left[\\log_{}x\\times \\log_{}\\left(\\log_{}x\\right)\\right]}{x}}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\log_{}x.\\diff{\\left[\\log_{}\\left(\\log_{}x\\right)\\right]}{x}+\\log_{}\\left(\\log_{}x\\right).\\diff{\\left(\\log_{}x\\right)}{x}}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\log_{}x\\times\\frac{1}{\\log_{}x}\\times\\frac{1}{x}+\\log_{}\\left(\\log_{}x\\right)\\times\\frac{1}{x}}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\frac{1}{x}+\\log_{}\\left(\\log_{}x\\right)\\times\\frac{1}{x}}\t\n\\end{align*}","id":"1-0-1-1-1-1-0-1","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1599045726760,"cs":"Gdk/mvfUbEYlHrNIzLrVIQ==","size":{"width":344,"height":160}}

{"code":"\\begin{align*}\n{\\diff{y}{x}}&={y\\left(\\frac{1}{x}+\\log_{}\\left(\\log_{}x\\right)\\times\\frac{1}{x}\\right)}\\\\\n{\\diff{y}{x}}&={\\left(\\log_{}x\\right)^{\\log_{}x}.\\left(\\frac{1}{x}+\\log_{}\\left(\\log_{}x\\right)\\times\\frac{1}{x}\\right)}\t\n\\end{align*}","type":"align*","id":"5-1-0-0","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1599045929434,"cs":"1LMJ0tq99FePe82QMCdMgw==","size":{"width":281,"height":80}}

8. cos (acos x + bsin x) , for some constant a and b

Solun:- Let y = cos (acos x + bsin x)

Differentiate w.r.t. x:-

{"type":"align*","id":"1-0-1-1-1-1-0-2","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{\\left[\\cos\\left(a\\cos x+b\\sin x\\right)\\right]}{x}}\\\\\n{\\diff{y}{x}}&={-\\sin\\left(a\\cos x+b\\sin x\\right)\\times\\left(-a\\sin x+b\\cos x\\right)}\\\\\n{\\diff{y}{x}}&={\\sin\\left(a\\cos x+b\\sin x\\right)\\times\\left(a\\sin x-b\\cos x\\right)}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1599046706405,"cs":"mQL+LfVotkQBP45eIiqfdQ==","size":{"width":344,"height":108}}

9. (sin x - cos x)(sin x - cos x)

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

Taking log both sides:-

⇒ log y = log (sin x - cos x)(sin x - cos x)

We know that log (x)n = nlog x

⇒ log y = (sin x - cos x).(log (sin x - cos x))

Differentiate w.r.t. x:-

{"id":"1-0-1-1-1-1-0-3","code":"\\begin{align*}\n{\\frac{1}{y}\\diff{y}{x}}&={\\diff{\\left[\\left(\\sin x-\\cos x\\right)\\times \\log_{}\\left(\\sin x-\\cos x\\right)\\right]}{x}}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\left(\\sin x-\\cos x\\right).\\diff{\\left[\\log_{}\\left(\\sin x-\\cos x\\right)\\right]}{x}+\\log_{}\\left(\\sin x-\\cos x\\right).\\diff{\\left(\\left(\\sin x-\\cos x\\right)\\right)}{x}}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\left(\\sin x-\\cos x\\right)\\times\\frac{\\left(\\cos x+\\sin x\\right)}{\\left(\\sin x-\\cos x\\right)}+\\log_{}\\left(\\sin x-\\cos x\\right).\\left(\\cos x+\\sin x\\right)}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\left(\\cos x+\\sin x\\right)+\\log_{}\\left(\\sin x-\\cos x\\right).\\left(\\cos x+\\sin x\\right)}\\\\\n{\\frac{1}{y}\\diff{y}{x}}&={\\left(\\cos x+\\sin x\\right)\\left[1+\\log_{}\\left(\\sin x-\\cos x\\right)\\right]}\t\n\\end{align*}","font":{"size":8,"color":"#000000","family":"Arial"},"type":"align*","ts":1599050798587,"cs":"seqh2yHsud2Z+HEx/RYmLg==","size":{"width":489,"height":173}}

{"type":"align*","id":"5-1-1","font":{"family":"Arial","color":"#000000","size":8},"code":"\\begin{align*}\n{\\diff{y}{x}}&={y\\left(\\sin x+\\cos x\\right).\\left(1+\\log_{}\\left(\\sin x-\\cos x\\right)\\right)}\\\\\n{\\diff{y}{x}}&={\\left(\\sin x-\\cos x\\right)^{\\left(\\sin x-\\cos x\\right)}.\\left(\\sin x+\\cos x\\right).\\left(1+\\log_{}\\left(\\sin x-\\cos x\\right)\\right)}\t\n\\end{align*}","ts":1599399842498,"cs":"aqHNJ3FOgUqjigfjgqlDSA==","size":{"width":404,"height":60}} (sinx > cosx)

10.  xx + xa + ax + aa , for some fixed a>0 and x>0

Solun:- Let y = xx + xa + ax + aa

And let u =  xx 

⇒ y = u + xa + ax + aa …....(1)

Taking log both sides:-

⇒ log u = log xx

We know that log (x)n = nlog x

⇒ log u = x.(log x)

Differentiate w.r.t. x:-

{"id":"1-0-1-1-1-1-0-4-0","type":"align*","font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\frac{1}{u}\\diff{u}{x}}&={\\diff{\\left[x\\times\\log_{}x\\right]}{x}}\\\\\n{\\frac{1}{u}\\diff{u}{x}}&={x.\\diff{\\left[\\left(\\log_{}x\\right)\\right]}{x}+\\left(\\log_{}x\\right).\\diff{\\left(x\\right)}{x}}\\\\\n{\\frac{1}{u}\\diff{u}{x}}&={x\\times\\frac{1}{x}+\\left(\\log_{}x\\right)\\times1}\\\\\n{\\frac{1}{u}\\diff{u}{x}}&={1+\\left(\\log_{}x\\right)}\t\n\\end{align*}","ts":1599052172421,"cs":"i4PxUPSLq5aqTBqIHgFs1Q==","size":{"width":256,"height":148}}

{"type":"align*","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\diff{u}{x}}&={u\\left(1+\\log_{}x\\right)}\\\\\n{\\diff{u}{x}}&={x^{x}.\\left(1+\\log_{}x\\right)}\t\n\\end{align*}","id":"5-1-0-1-0-0","ts":1599052271198,"cs":"1TFKEuLOaLIcWdicmiCtWQ==","size":{"width":141,"height":69}}

From Eq. 1:-

Differentiate w.r.t. x:-

We know that:-

{"code":"\\begin{align*}\n{\\diff{\\left(x^{n}\\right)}{x}}&={nx^{n-1}}\\\\\n{\\diff{\\left(a^{x}\\right)}{x}}&={a^{x}\\log_{}a}\\\\\n{\\diff{\\left(a^{a}\\right)}{x}}&={0}\t\n\\end{align*}","type":"align*","id":"6","font":{"family":"Arial","color":"#000000","size":10},"ts":1599052743979,"cs":"pFdmgNjr1EPfgHePZQPI6w==","size":{"width":116,"height":112}}

{"font":{"color":"#000000","family":"Arial","size":10},"id":"1-0-1-1-1-1-0-5","type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{u}{x}+\\diff{\\left(x^{a}\\right)}{x}+\\diff{\\left(a^{x}\\right)}{x}+\\diff{\\left(a^{a}\\right)}{x}}\\\\\n{\\diff{y}{x}}&={x^{x}\\left(1+\\log_{}x\\right)+ax^{a-1}+a^{x}\\log_{e}a+0}\\\\\n{\\diff{y}{x}}&={x^{x}\\left(1+\\log_{}x\\right)+ax^{a-1}+a^{x}\\log_{e}a}\t\n\\end{align*}","ts":1599052614173,"cs":"PIMAeW7xUna0GZovDW2R1g==","size":{"width":293,"height":108}}

{"type":"$","id":"7","code":"$11.\\,x^{x^{2}-3}+\\left(x-3\\right)^{x^{2}},for\\,x>3$","font":{"color":null,"size":12,"family":"Arial"},"ts":1599297866843,"cs":"cNsdG1Cc6TUEkHqByiJGyg==","size":{"width":268,"height":26}}

Solun:- Let 

{"type":"$","code":"$y=x^{x^{2}-3}+\\left(x-3\\right)^{x^{2}}$","id":"8","font":{"family":"Arial","color":"#000000","size":12},"ts":1599297897132,"cs":"8aVER19Pdb27ndgmwJLI6Q==","size":{"width":184,"height":26}}

And let 

{"font":{"size":12,"family":"Arial","color":"#000000"},"type":"$","code":"$u=x^{x^{2}-3}$","id":"9","ts":1599298017912,"cs":"UPhxhW5zPVbg9sMxRtKGvQ==","size":{"width":84,"height":20}}

{"type":"$","id":"10-0","font":{"size":12,"color":"#000000","family":"Arial"},"code":"$v=\\left(x-3\\right)^{x^{2}}$","ts":1599298085276,"cs":"LcJ3txIBU5ExPnAiah5Bmw==","size":{"width":114,"height":26}}

⇒ y = u + v …....(1)

{"font":{"size":12,"family":"Arial","color":"#000000"},"type":"$","code":"$u=x^{x^{2}-3}$","id":"9","ts":1599298017912,"cs":"UPhxhW5zPVbg9sMxRtKGvQ==","size":{"width":84,"height":20}}

Taking log both sides:-

We know that log (x)n = nlog x

⇒ log u =(x2-3).log x

Differentiate w.r.t. x:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\frac{1}{u}\\diff{u}{x}}&={\\diff{\\left[\\left(x^{2}-3\\right)\\times\\log_{}x\\right]}{x}}\\\\\n{\\frac{1}{u}\\diff{u}{x}}&={\\left(x^{2}-3\\right).\\diff{\\left[\\left(\\log_{}x\\right)\\right]}{x}+\\left(\\log_{}x\\right).\\diff{\\left(x^{2}-3\\right)}{x}}\\\\\n{\\frac{1}{u}\\diff{u}{x}}&={\\left(x^{2}-3\\right)\\times\\frac{1}{x}+\\left(\\log_{}x\\right)\\times2x}\\\\\n{\\frac{1}{u}\\diff{u}{x}}&={\\frac{\\left(x^{2}-3\\right)}{x}+2x\\left(\\log_{}x\\right)}\t\n\\end{align*}","type":"align*","id":"1-0-1-1-1-1-0-6-0","ts":1599299122344,"cs":"6mEUYq29eNoY5Px/EtIFNw==","size":{"width":340,"height":158}}

{"id":"5-1-0-1-1-0","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","code":"\\begin{align*}\n{\\diff{u}{x}}&={u\\left[\\frac{\\left(x^{2}-3\\right)}{x}+2x\\log_{}x\\right]}\\\\\n{\\diff{u}{x}}&={x^{x^{2}-3}.\\left[\\frac{\\left(x^{2}-3\\right)}{x}+2x\\log_{}x\\right]}\t\n\\end{align*}","ts":1599299192948,"cs":"7XyzRtJ5ZLB1u9gj2R34OA==","size":{"width":241,"height":100}}

{"type":"$","id":"10-1","font":{"size":12,"color":"#000000","family":"Arial"},"code":"$v=\\left(x-3\\right)^{x^{2}}$","ts":1599298085276,"cs":"4/awrGdKn6VN47OnkjdQRw==","size":{"width":114,"height":26}}

Taking log both sides:-

We know that log (x)n = nlog x

⇒ log u =(x2).log (x-3)

Differentiate w.r.t. x:-

{"id":"1-0-1-1-1-1-0-6-1","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\frac{1}{v}\\diff{v}{x}}&={\\diff{\\left[x^{2}\\times\\log_{}\\left(x-3\\right)\\right]}{x}}\\\\\n{\\frac{1}{v}\\diff{v}{x}}&={x^{2}.\\diff{\\left[\\log_{}\\left(x-3\\right)\\right]}{x}+\\left(\\log_{}\\left(x-3\\right)\\right).\\diff{\\left(x^{2}\\right)}{x}}\\\\\n{\\frac{1}{v}\\diff{v}{x}}&={\\frac{x^{2}}{x-3}+2x.\\left(\\log_{}\\left(x-3\\right)\\right)}\\\\\n{\\frac{1}{v}\\diff{v}{x}}&={\\frac{x^{2}}{x-3}+2x\\left(\\log_{}\\left(x-3\\right)\\right)}\t\n\\end{align*}","type":"align*","ts":1599400396718,"cs":"D8fg6dL/fxFGGHuM0Hxgiw==","size":{"width":332,"height":160}}

{"id":"5-1-0-1-1-1","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\diff{v}{x}}&={u\\left[\\frac{x^{2}}{x-3}+2x\\left(\\log_{}\\left(x-3\\right)\\right)\\right]}\\\\\n{\\diff{v}{x}}&={\\left(x-3\\right)^{x^{2}}.\\left[\\frac{x^{2}}{x-3}+2x\\log_{}\\left(x-3\\right)\\right]}\t\n\\end{align*}","type":"align*","ts":1599400486422,"cs":"dkXLiMTDgoGULd/fuHQe0g==","size":{"width":281,"height":82}}

{"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{u}{x}+\\diff{\\left(v\\right)}{x}}\\\\\n{\\diff{y}{x}}&={x^{x^{2}-3}.\\left[\\frac{\\left(x^{2}-3\\right)}{x}+2x\\log_{}x\\right]+\\left(x-3\\right)^{x^{2}}.\\left[\\frac{x^{2}}{x-3}+2x\\log_{}\\left(x-3\\right)\\right]}\t\n\\end{align*}","font":{"family":"Arial","size":8,"color":"#000000"},"id":"1-0-1-1-1-1-0-7","type":"align*","ts":1599400502792,"cs":"JbgUFqc2gRD0VatUqb/jxg==","size":{"width":426,"height":73}}

12.  Find dy/dx, if y = 12(1 - cos t), x = 10(t - sin t), -Ï€/2 < t < Ï€/2

Solun:- Given y = 12(1 - cos t) and x = 10(t - sin t)

y = 12(1 - cos t)

Differentiate w.r.t. x:-

{"type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{\\left[12\\left(1-\\cos t\\right)\\right]}{x}}\\\\\n{\\diff{y}{x}}&={12\\left(0+\\sin t\\right)\\diff{t}{x}}\\\\\n{\\diff{y}{x}}&={12\\sin t\\diff{t}{x}.....\\left(1\\right)}\t\n\\end{align*}","id":"1-0-1-1-1-1-0-4-1-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1599400648098,"cs":"7JrNImcZ/0I0DACwm/vrUA==","size":{"width":172,"height":108}}

x = 10(t - sin t)

{"id":"5-1-0-1-0-1","font":{"family":"Arial","color":"#000000","size":10},"type":"align*","code":"\\begin{align*}\n{\\diff{x}{x}}&={\\diff{\\left[10\\left(t-\\sin t\\right)\\right]}{x}}\\\\\n{\\diff{x}{x}}&={10\\left(1-\\cos t\\right)\\diff{t}{x}}\\\\\n{1}&={10\\left(1-\\cos t\\right)\\diff{t}{x}.....\\left(2\\right)}\t\n\\end{align*}","ts":1599301547846,"cs":"ClTUR6k2AgeO8a6Qkl9MQw==","size":{"width":210,"height":108}}

Divide 1/2:-

{"id":"11","type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{12\\sin t}{10\\left(1-\\cos t\\right)}}\\\\\n{\\diff{y}{x}}&={\\frac{12\\times2\\sin\\frac{t}{2}.\\cos\\frac{t}{2}}{10\\times2\\sin^{2}\\frac{t}{2}}}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"ts":1599301953629,"cs":"fvzUpJV3HoZx5kYFFs8rpA==","size":{"width":176,"height":85}}

{"id":"12","type":"$","code":"$\\diff{y}{x}=\\frac{6}{5}\\cot\\frac{t}{2}$","font":{"family":"Arial","color":"#000000","size":12},"ts":1599302025849,"cs":"EC4yD4qAn1RpAVj7UJD0ZQ==","size":{"width":114,"height":26}}

{"font":{"family":"Arial","size":12,"color":"#434343"},"code":"$13. Find \\,\\diff{y}{x}, if \\,y = \\sin ^{-1}x + \\sin^{-1}{\\sqrt []{1-x^{2}}}, \\,-1\\leq x\\leqslant1$","type":"$","id":"13","ts":1599303908426,"cs":"wUf2nLCGsbYONd9Ap4cynw==","size":{"width":496,"height":26}}

Solun:- Given y = sin-1x + sin-1 √1-x2

Differentiate w.r.t. x:-

{"id":"1-0-1-1-1-1-0-4-1-1","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{\\left[\\sin^{-1}x+\\sin^{-1}{\\sqrt[]{1-x^{2}}}\\right]}{x}}\\\\\n{\\diff{y}{x}}&={\\frac{1}{{\\sqrt[]{1-x^{2}}}}+\\frac{-2x}{2{\\sqrt[]{1-x^{2}}}{\\sqrt[]{1-\\left({\\sqrt[]{1-x^{2}}}\\right)^{2}}}}}\\\\\n{\\diff{y}{x}}&={\\frac{1}{{\\sqrt[]{1-x^{2}}}}-\\frac{x}{{\\sqrt[]{1-x^{2}}}{\\sqrt[]{x^{2}}}}}\\\\\n{\\diff{y}{x}}&={\\frac{1}{{\\sqrt[]{1-x^{2}}}}-\\frac{1}{{\\sqrt[]{1-x^{2}}}}}\\\\\n{\\diff{y}{x}}&={0}\t\n\\end{align*}","type":"align*","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1599304370626,"cs":"3gs7fuzDhQKffSHSJGZT4A==","size":{"width":324,"height":234}}

14. If {"id":"15","type":"$","font":{"color":"#222222","family":"Arial","size":12},"code":"$x{\\sqrt []{1+y }}+ y{\\sqrt []{1+x }}= 0$","ts":1599305330091,"cs":"HA3rUQTzB984WftP4FCJLg==","size":{"width":213,"height":25}}, for , -1 < x < 1, prove that

{"id":"14-0","code":"$\\diff{y}{x}=\\frac{-1}{\\left(1+x^{2}\\right)}$","type":"$","font":{"color":"#000000","size":12,"family":"Arial"},"ts":1599305179331,"cs":"GTBVQK753RWbJQO2EwfpAw==","size":{"width":104,"height":30}}

Solun:- Given 

{"id":"16","font":{"color":"#000000","size":10,"family":"Arial"},"type":"$","code":"$x{\\sqrt []{1+y }}+ y{\\sqrt []{1+x }}= 0$","ts":1599305448824,"cs":"kAfOYNLvP+jV0TixfM7I6w==","size":{"width":165,"height":20}}

{"font":{"family":"Arial","color":"#000000","size":10},"id":"19-0","code":"\\begin{align*}\n{x{\\sqrt[]{1+y}}}&={-y{\\sqrt[]{1+x}}}\\\\\n{x^{2}\\left(1+y\\right)}&={y^{2}\\left(1+x\\right)}\\\\\n{x^{2}+x^{2}y}&={y^{2}+y^{2}x}\\\\\n{x^{2}-y^{2}}&={y^{2}x-x^{2}y}\t\n\\end{align*}","type":"align*","ts":1599306721289,"cs":"QKmUnWRJ5o0mFlHFmk+psA==","size":{"width":153,"height":88}}

(x-y)(x+y) = xy(y-x)

(x+y) = -xy

x+y-xy = 0

y(1-x) = -x

{"type":"$","id":"21-0","font":{"color":"#000000","family":"Arial","size":12},"code":"$y=\\frac{-x}{1-x}$","ts":1599307000294,"cs":"L+hJOuaRf124xz7HtkCrww==","size":{"width":72,"height":24}}

Differentiate w.r.t. x:-

{"code":"\\begin{align*}\n{\\diff{\\left(y\\right)}{x}}&={\\diff{\\left(\\frac{-x}{1-x}\\right)}{x}}\\\\\n{\\diff{y}{x}}&={\\frac{\\left(1-x\\right)\\diff{\\left(-x\\right)}{x}-\\left(-x\\right)\\diff{\\left(1-x\\right)}{x}}{\\left(1-x\\right)^{2}}}\\\\\n{\\diff{y}{x}}&={\\frac{-\\left(1-x\\right)-x}{\\left(1-x\\right)^{2}}}\\\\\n{\\diff{y}{x}}&={\\frac{-1}{\\left(1-x\\right)^{2}}}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"17","ts":1599307483311,"cs":"lPea4OAfUb1pZkWuqnDFkQ==","size":{"width":244,"height":181}}

{"code":"\\begin{align*}\n{\\diff{\\left(y\\right)}{x}}&={\\diff{\\left(\\frac{-x}{1-x}\\right)}{x}}\\\\\n{\\diff{y}{x}}&={\\frac{\\left(1-x\\right)\\diff{\\left(-x\\right)}{x}-\\left(-x\\right)\\diff{\\left(1-x\\right)}{x}}{\\left(1-x\\right)^{2}}}\\\\\n{\\diff{y}{x}}&={\\frac{-\\left(1-x\\right)-x}{\\left(1-x\\right)^{2}}}\\\\\n{\\diff{y}{x}}&={\\frac{-1}{\\left(1-x\\right)^{2}}}\t\n\\end{align*}","font":{"color":"#000000","size":10,"family":"Arial"},"type":"align*","id":"17","ts":1599307483311,"cs":"lPea4OAfUb1pZkWuqnDFkQ==","size":{"width":244,"height":181}}

15. If (x - a)2 + (y - b)2 = c2 , for some c>0, prove that

{"font":{"size":14,"color":"#000000","family":"Arial"},"type":"$","id":"14-1-0","code":"$\\frac{\\left[1+\\left(\\diff{y}{x}\\right)^{2}\\right]^{\\frac{3}{2}}}{\\frac{d^{2}y}{dx^{2}}}$","ts":1599310432808,"cs":"qaD/YQBtZf/Qk4dd0p5O+w==","size":{"width":104,"height":76}}is constant independent of a and b.

Solun:- Given (x - a)2 + (y - b)2 = c2

{"type":"align*","id":"19-1-0","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\diff{\\left[\\left(x-a\\right)^{2}+\\left(y-b\\right)^{2}\\right]}{x}}&={\\diff{\\left(c^{2}\\right)}{x}}\\\\\n{2\\left(x-a\\right)+2\\left(y-b\\right)\\diff{y}{x}}&={0}\t\n\\end{align*}","ts":1599310949428,"cs":"m6a0bUIMDhGQUVzM9+WD5w==","size":{"width":222,"height":84}}

{"font":{"color":"#000000","family":"Arial","size":12},"id":"21-1-0","type":"$","code":"$\\diff{y}{x}=\\frac{-\\left(x-a\\right)}{\\left(y-b\\right)}$","ts":1599311033122,"cs":"TZz5Qf3pLhfYq465t6TbZQ==","size":{"width":110,"height":32}}

Differentiate w.r.t. x:-

{"type":"align*","id":"22-0","font":{"family":"Arial","size":10,"color":"#000000"},"code":"\\begin{align*}\n{\\frac{d^{2}y}{dx^{2}}}&={\\diff{\\left[\\frac{-\\left(x-a\\right)}{y-b}\\right]}{x}}\\\\\n{\\frac{d^{2}y}{dx^{2}}}&={\\frac{\\left(y-b\\right)\\diff{\\left[-\\left(x-a\\right)\\right]}{x}+\\left(x-a\\right)\\diff{\\left(y-b\\right)}{x}}{\\left(y-b\\right)^{2}}}\\\\\n{\\frac{d^{2}y}{dx^{2}}}&={\\frac{-\\left(y-b\\right)+\\left(x-a\\right)\\diff{y}{x}}{\\left(y-b\\right)^{2}}}\\\\\n{\\frac{d^{2}y}{dx^{2}}}&={\\frac{-\\left(y-b\\right)+\\left(x-a\\right)\\left\\{\\frac{-\\left(x-a\\right)}{y-b}\\right\\}}{\\left(y-b\\right)^{3}}}\\\\\n{\\frac{d^{2}y}{dx^{2}}}&={\\frac{-\\left(y-b\\right)^{2}-\\left(x-a\\right)^{2}}{\\left(y-b\\right)^{3}}}\\\\\n{\\frac{d^{2}y}{dx^{2}}}&={-\\left[\\frac{\\left(y-b\\right)^{2}+\\left(x-a\\right)^{2}}{\\left(y-b\\right)^{3}}\\right]}\t\n\\end{align*}","ts":1599401738971,"cs":"A567Le8QfiQT6Bq9C82CCA==","size":{"width":272,"height":312}}

Taking L.H.S:-

{"id":"23","type":"$","code":"$=\\frac{\\left[1+\\left(\\diff{y}{x}\\right)^{2}\\right]^{\\frac{3}{2}}}{\\frac{d^{2}y}{dx^{2}}}$","font":{"family":"Arial","size":14,"color":"#000000"},"ts":1599311604462,"cs":"TuptOOLE7Rsk5/OWZSPbBQ==","size":{"width":128,"height":76}}

{"type":"align*","code":"\\begin{align*}\n{}&={\\frac{\\left[1+\\left[\\frac{\\left(x-a\\right)}{\\left(y-b\\right)}\\right]^{2}\\right]^{\\frac{3}{2}}}{-\\left[\\frac{\\left(y-b\\right)^{2}+\\left(x-a\\right)^{2}}{\\left(y-b\\right)^{3}}\\right]}}\\\\\n{}&={-\\,\\frac{\\left[\\frac{\\left(y-b\\right)^{2}+\\left(x-a\\right)^{2}}{\\left(y-b\\right)^{2}}\\right]^{\\frac{3}{2}}}{\\left[\\frac{\\left(y-b\\right)^{2}+\\left(x-a\\right)^{2}}{\\left(y-b\\right)^{3}}\\right]}}\\\\\n{}&={-\\,\\frac{\\left[\\frac{c^{2}}{\\left(y-b\\right)^{2}}\\right]^{\\frac{3}{2}}}{\\left[\\frac{c^{2}}{\\left(y-b\\right)^{3}}\\right]}}\\\\\n{}&={-\\,\\frac{\\left[\\frac{c^{3}}{\\left(y-b\\right)^{3}}\\right]}{\\left[\\frac{c^{2}}{\\left(y-b\\right)^{3}}\\right]}}\\\\\n{}&={-c}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":11},"id":"24","ts":1599312022009,"cs":"5LPO/jqSs49g7Woo7FTUbw==","size":{"width":170,"height":366}}

Hence L.H.S is independent of a and b.

16. If cos y = xcos (a+y), with cos a ≠ ±1, prove that

{"type":"$","code":"$\\diff{y}{x}=\\frac{\\cos^{2}\\left(a+y\\right)}{\\sin a}$","id":"14-1-1","font":{"color":"#000000","size":14,"family":"Arial"},"ts":1599386780576,"cs":"TQLfySm0DHPFKxpOlkAvxg==","size":{"width":136,"height":32}}

Solun:- Given cos y = xcos (a+y)

Differentiate w.r.t. x:-

{"font":{"family":"Arial","color":"#000000","size":10},"code":"\\begin{align*}\n{\\diff{\\left[cos y \\right]}{x}}&={\\diff{\\left(xcos (a+y)\\right)}{x}}\\\\\n{-\\sin y.\\diff{y}{x}}&={x\\diff{\\left(\\cos\\left(a+y\\right)\\right)}{x}+\\cos\\left(a+y\\right).\\diff{x}{x}}\\\\\n{-\\sin y.\\diff{y}{x}}&={-x.\\sin\\left(a+y\\right)\\diff{y}{x}+\\cos\\left(a+y\\right)}\t\n\\end{align*}","type":"align*","id":"19-1-1","ts":1599386996085,"cs":"K/9YSkH/+rOUN1aKtSkEPg==","size":{"width":324,"height":110}}

{"font":{"size":10,"family":"Arial","color":"#000000"},"type":"$","code":"$\\diff{y}{x}\\left(x\\sin\\left(a+y\\right)-\\sin y\\right)=\\cos\\left(a+y\\right)$","id":"21-1-1-0","ts":1599387103512,"cs":"OyflHo8jVGEICI12TTo7mg==","size":{"width":249,"height":20}}

Given x = cosy/cos(a+y)

{"type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}\\left(\\frac{\\cos y}{\\cos\\left(a+y\\right)}.\\sin\\left(a+y\\right)-\\sin y\\right)}&={\\cos\\left(a+y\\right)}\\\\\n{\\diff{y}{x}\\left(\\cos y.\\sin\\left(a+y\\right)-\\cos\\left(a+y\\right).\\sin y\\right)}&={\\cos^{2}\\left(a+y\\right)}\t\n\\end{align*}","id":"25-0","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1599387240101,"cs":"OcVRfj5fOlPnlZbxmzBAow==","size":{"width":364,"height":76}}

We know that sinA.cos B-cos A.sinB = sin(A - B)

{"font":{"size":10,"family":"Arial","color":"#000000"},"id":"22-1-0","type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{\\cos^{2}\\left(a+y\\right)}{\\sin a}}\t\n\\end{align*}","ts":1599387634917,"cs":"pjYvbXoACdmuu4NnWqNYzA==","size":{"width":128,"height":34}}

17. If x = a(cos t + tsin t) and y = a(sin t - tcos t),find

{"type":"$","id":"14-1-2","code":"$\\frac{d^{2}y}{dx^{2}}.$","font":{"family":"Arial","color":"#000000","size":14},"ts":1599387818479,"cs":"ocrBMmFzhRxd14PQobuBFA==","size":{"width":40,"height":32}}

Solun:- Given x = a(cos t + tsin t)

Differentiate w.r.t. x:-

{"type":"align*","id":"19-1-2","code":"\\begin{align*}\n{\\diff{x}{x}}&={\\diff{\\left[a\\left(\\cos t+t.\\sin t\\right)\\right]}{x}}\\\\\n{1}&={a\\left[-\\sin t\\diff{t}{x}+\\diff{\\left(t.\\sin t\\right)}{x}\\right]}\\\\\n{1}&={a\\left[-\\sin t\\diff{t}{x}+\\left(t\\cos t+\\sin t\\right)\\diff{t}{x}\\right]}\\\\\n{1}&={a\\left[-\\sin t+\\left(t\\cos t+\\sin t\\right)\\right]\\diff{t}{x}}\\\\\n{1}&={a\\left[t\\cos t\\right]\\diff{t}{x}.....\\left(1\\right)}\t\n\\end{align*}","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1599388272356,"cs":"OTRlWQkU6FuwT/5aywErBw==","size":{"width":280,"height":194}}

Given y = a(sin t - tcos t)

Differentiate w.r.t. x:-

{"font":{"color":"#000000","size":10,"family":"Arial"},"id":"21-1-1-1-0","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{\\left[a\\left(\\sin t-t.\\cos t\\right)\\right]}{x}}\\\\\n{\\diff{y}{x}}&={a\\left(\\cos t\\diff{t}{x}-\\diff{\\left(t.\\cos t\\right)}{x}\\right)}\\\\\n{\\diff{y}{x}}&={a\\left[\\cos t\\diff{t}{x}-\\diff{\\left(t.\\cos t\\right)}{x}\\right]}\\\\\n{\\diff{y}{x}}&={a\\left[\\cos t\\diff{t}{x}-\\left(-t.\\sin t+\\cos t\\right)\\diff{t}{x}\\right]}\\\\\n{\\diff{y}{x}}&={a\\left[\\cos t-\\left(-t.\\sin t+\\cos t\\right)\\right]\\diff{t}{x}}\\\\\n{\\diff{y}{x}}&={a\\left[t.\\sin t\\right]\\diff{t}{x}.....\\left(2\\right)}\t\n\\end{align*}","type":"align*","ts":1599389164922,"cs":"OrE2IGIXHOMmGy94rhbxeg==","size":{"width":284,"height":237}}

Divide Eq. 2/1:-

{"font":{"size":10,"family":"Arial","color":"#000000"},"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{a\\left(t.\\sin t\\right)}{a\\left(t.\\cos t\\right)}}\\\\\n{\\diff{y}{x}}&={\\tan t}\t\n\\end{align*}","id":"25-1","type":"align*","ts":1599389252320,"cs":"ZkcHj01+Q90B2ZnNhmn55w==","size":{"width":116,"height":76}}

Again differentiate w.r.t. x:-

{"type":"align*","code":"\\begin{align*}\n{\\frac{d^{2}y}{dx^{2}}}&={\\sec^{2}t.\\diff{t}{x}}\\\\\n{\\frac{d^{2}y}{dx^{2}}}&={\\sec^{2}t\\times\\frac{1}{a\\left(t.\\cos t\\right)}\\left(From\\,Eq.\\,1\\right)}\\\\\n{\\frac{d^{2}y}{dx^{2}}}&={\\frac{\\sec^{3}t}{at}}\t\n\\end{align*}","font":{"family":"Arial","color":"#000000","size":10},"id":"22-1-1","ts":1599403987168,"cs":"EMw5uzjFoRmkC1UeFbB4Fg==","size":{"width":268,"height":120}}

18. If f(x) = |x|3 , show that f”(x) exists for all real x and find it.

Solun:- Given f(x) = |x|3

We know that

{"font":{"color":"#000000","family":"Arial","size":10},"type":"align*","code":"\\begin{align*}\n{\\left|x\\right|}&={\\begin{cases}\n{x,}&{x\\geq0}\\\\\n{-x}&{x<0}\\\\\n\\end{cases}}\\\\\n{\\left|x\\right|^{3}}&={\\begin{cases}\n{x^{3},}&{x\\geq0}\\\\\n{-x^{3},}&{x<0}\\\\\n\\end{cases}}\\\\\n{f\\left(x\\right)}&={\\begin{cases}\n{x^{3},}&{x\\geq0}\\\\\n{-x^{3},}&{x<0}\\\\\n\\end{cases}}\t\n\\end{align*}","id":"26","ts":1599390084238,"cs":"gmxLxfLssjrPUChT+y7NHA==","size":{"width":148,"height":122}}

Differentiate w.r.t. x:-

{"font":{"family":"Arial","color":"#000000","size":10},"id":"19-1-2","code":"\\begin{align*}\n{f^{\\prime }\\left(x\\right)}&={\\begin{cases}\n{3x^{2},}&{x\\geq0}\\\\\n{-3x^{2}}&{x<0}\\\\\n\\end{cases}}\t\n\\end{align*}","type":"align*","ts":1599390244063,"cs":"EloO3pW4kY++X8B6ex+BeQ==","size":{"width":156,"height":37}}

Again differentiate w.r.t. x:-

{"type":"align*","id":"21-1-1-1-1","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{f^{\\prime \\prime }\\left(x\\right)}&={\\begin{cases}\n{6x,}&{x\\geq0}\\\\\n{-6x}&{x<0}\\\\\n\\end{cases}}\t\n\\end{align*}","ts":1599390403005,"cs":"C2kWMQFpXNo7aiQufegLdg==","size":{"width":153,"height":37}}

Thus f”(x) exists.

19. Using mathematical induction prove that {"id":"27-0","type":"$","font":{"family":"Arial","color":"#000000","size":11},"code":"$\\diff{\\left(x^{n}\\right)}{x}=nx^{n-1}$","ts":1599390577419,"cs":"x6XurCtdSAcWdQHaWKHZmQ==","size":{"width":108,"height":24}} for all positive integers n.

Solun:- Let P(n):{"id":"27-1","type":"$","font":{"family":"Arial","color":"#000000","size":11},"code":"$\\diff{\\left(x^{n}\\right)}{x}=nx^{n-1}$","ts":1599390577419,"cs":"GiofnvbWoa8BYSnPuTiSBA==","size":{"width":108,"height":24}}

Put n=1

P(1):

{"code":"\\begin{align*}\n{\\diff{\\left(x\\right)}{x}}&={x^{0}}\\\\\n{1}&={1}\t\n\\end{align*}","font":{"color":"#000000","family":"Arial","size":10},"id":"28","type":"align*","ts":1599390686416,"cs":"wUbcA/784Cy/WC4f0U7yUA==","size":{"width":73,"height":54}}(True)

Hence P(k) is true.

Put n = k, P(k):

{"id":"29-0","type":"$","font":{"size":11,"color":"#000000","family":"Arial"},"code":"$\\diff{\\left(x^{k}\\right)}{x}=kx^{k-1}....\\left(1\\right)$","ts":1599390976807,"cs":"asXsRnSz630aPPbXIh0rvQ==","size":{"width":162,"height":28}}

Put n=k+1

Prove that P(k+1):

{"type":"align*","id":"29-1-0","font":{"color":"#000000","family":"Arial","size":10},"code":"\\begin{align*}\n{\\diff{\\left(x^{k+1}\\right)}{x}}&={\\left(k+1\\right)x^{k}}\\\\\n{\\diff{\\left(x^{k}\\times x\\right)}{x}}&={\\left(k+1\\right)x^{k}}\t\n\\end{align*}","ts":1599391222000,"cs":"basvOHxQ1MCVX4bkYe41aQ==","size":{"width":158,"height":40}}

Taking L.H.S:-

{"id":"33","type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{}&={\\diff{\\left(x^{k+1}\\right)}{x}}\\\\\n{}&={\\diff{\\left(x^{k}.x\\right)}{x}}\t\n\\end{align*}","ts":1599404242771,"cs":"nuRHI7wpoJ0E3DY5eCcJ5w==","size":{"width":82,"height":80}}

We know that

{"id":"32","font":{"color":"#000000","family":"Arial","size":12},"code":"$\\diff{\\left(u.v\\right)}{x}=u\\diff{\\left(v\\right)}{x}+v\\diff{\\left(u\\right)}{x}$","type":"$","ts":1599392268877,"cs":"ug42mET6ecr8kxQOxd1mDg==","size":{"width":192,"height":28}}

{"code":"\\begin{align*}\n{}&={x^{k}\\diff{x}{x}+x\\diff{\\left(x^{k}\\right)}{x}}\\\\\n{}&={x^{k}+x\\times\\left(kx^{k-1}\\right)\\left(From\\,Eq.\\,1\\right)}\\\\\n{}&={x^{k}+\\left(kx^{k}\\right)}\\\\\n{}&={x^{k}\\left(k+1\\right)}\t\n\\end{align*}","id":"31","type":"align*","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1599391919869,"cs":"qyZMy6kkZUauvhMTk45I5A==","size":{"width":229,"height":109}}

= R.H.S

Hence Proved

20. Using the fact that sin(A+B) = sinA.cosB + cosAsinB and the differentiation, obtain the sum formula for cosines.

Solun:- Let sin(A+B) = sinA.cosB + cosAsinB

Differentiate w.r.t. x:-

{"id":"34","type":"align*","code":"\\begin{align*}\n{\\diff{\\left[\\sin\\left(A+B\\right)\\right]}{x}}&={\\diff{\\left(\\sin A\\cos B+\\cos A\\sin B\\right)}{x}}\\\\\n{\\cos\\left(A+B\\right)\\diff{\\left(A+B\\right)}{x}}&={-\\sin A\\sin B\\diff{B}{x}+\\cos A\\cos B\\diff{A}{x}+\\cos A\\cos B\\diff{B}{x}-\\sin A\\sin B\\diff{A}{x}}\\\\\n{\\cos\\left(A+B\\right)\\diff{\\left(A+B\\right)}{x}}&={\\diff{A}{x}\\left(\\cos A\\cos B-\\sin A\\sin B\\right)+\\diff{B}{x}\\left(\\cos A\\cos B-\\sin A\\sin B\\right)}\\\\\n{\\cos\\left(A+B\\right)\\diff{\\left(A+B\\right)}{x}}&={\\left(\\cos A\\cos B-\\sin A\\sin B\\right)\\left(\\diff{A}{x}+\\diff{B}{x}\\right)}\\\\\n{\\cos\\left(A+B\\right)\\diff{\\left(A+B\\right)}{x}}&={\\left(\\cos A\\cos B-\\sin A\\sin B\\right)\\left(\\diff{\\left(A+B\\right)}{x}\\right)}\t\n\\end{align*}","font":{"size":8,"color":"#000000","family":"Arial"},"ts":1599395408779,"cs":"jsuueGpEJup4XXwc65eKGQ==","size":{"width":550,"height":169}}

cos(A + B) = (cosA.cosB - sinA.sinB) (Hence Proved)

22. If{"font":{"color":"#000000","family":"Arial","size":10},"type":"$","id":"35-0","code":"$y=\\begin{vmatrix}\n{f\\left(x\\right)}&{g\\left(x\\right)}&{h\\left(x\\right)}\\\\\n{l}&{m}&{n}\\\\\n{a}&{b}&{c}\\\\\n\\end{vmatrix}$","ts":1599395685144,"cs":"ygjqs5JoI+8ByLM8elLgCA==","size":{"width":160,"height":60}},prove that{"font":{"size":10,"color":"#000000","family":"Arial"},"id":"36","code":"$\\diff{y}{x}=\\begin{vmatrix}\n{f^{\\prime }\\left(x\\right)}&{g^{\\prime }\\left(x\\right)}&{h^{\\prime }\\left(x\\right)}\\\\\n{l}&{m}&{n}\\\\\n{a}&{b}&{c}\\\\\n\\end{vmatrix}$","type":"$","ts":1599395846031,"cs":"CRVU74CO6e3ziMJxuGiUwg==","size":{"width":184,"height":60}}.

Solun:- Given 

{"font":{"color":"#000000","family":"Arial","size":10},"type":"$","id":"35-1","code":"$y=\\begin{vmatrix}\n{f\\left(x\\right)}&{g\\left(x\\right)}&{h\\left(x\\right)}\\\\\n{l}&{m}&{n}\\\\\n{a}&{b}&{c}\\\\\n\\end{vmatrix}$","ts":1599395685144,"cs":"ZuCOu0YzZ/Gms3qeiwXPTA==","size":{"width":160,"height":60}}

y = f(x)(mc-bn) - g(x)(lc-an) + h(x)(lb-am)

Differentiate w.r.t. x:-

{"type":"align*","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\left(mc-bn\\right)f^{\\prime }\\left(x\\right)-\\left(lc-an\\right)g^{\\prime }\\left(x\\right)+\\left(lb-am\\right)h^{\\prime }\\left(x\\right)}\t\n\\end{align*}","id":"37-0","font":{"size":10,"color":"#000000","family":"Arial"},"ts":1599396436414,"cs":"AokBMc16k48gmuaOhvraaw==","size":{"width":381,"height":32}}

{"code":"$\\diff{y}{x}=\\begin{vmatrix}\n{f^{\\prime }\\left(x\\right)}&{g^{\\prime }\\left(x\\right)}&{h^{\\prime }\\left(x\\right)}\\\\\n{l}&{m}&{n}\\\\\n{a}&{b}&{c}\\\\\n\\end{vmatrix}$","id":"38-0","type":"$","font":{"size":10,"family":"Arial","color":"#000000"},"ts":1599396505777,"cs":"08CJ08JGpM4le5z6MxtfPw==","size":{"width":184,"height":60}}

23. If {"id":"39-0","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$y=e^{a\\cos^{-1}x}$","type":"$","ts":1599396649181,"cs":"Mw7GERlLIkryo0dN8E4j/g==","size":{"width":78,"height":17}}, -1 ≤ x ≤ 1 show that 

{"font":{"color":"#000000","size":12,"family":"Arial"},"code":"$\\left(1-x^{2}\\right)\\frac{d^{2}y}{dx^{2}}-x\\diff{y}{x}-a^{2}y=0$","id":"40","type":"$","ts":1599396769662,"cs":"K4cgvO1vNrLEgYAH9Kk5VQ==","size":{"width":256,"height":29}}

Solun:- Given {"id":"39-1","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$y=e^{a\\cos^{-1}x}$","type":"$","ts":1599396649181,"cs":"OUFzHMgCUe2IBZuKaO2bqg==","size":{"width":78,"height":17}}

Differentiate w.r.t. x:-

{"code":"\\begin{align*}\n{\\diff{y}{x}}&={\\diff{\\left(e^{a\\cos^{-1}x}\\right)}{x}}\\\\\n{\\diff{y}{x}}&={e^{a\\cos^{-1}x}.\\diff{\\left(a\\cos^{-1}x\\right)}{x}}\\\\\n{\\diff{y}{x}}&={e^{a\\cos^{-1}x}.\\left(\\frac{-a}{{\\sqrt[]{1-x^{2}}}}\\right)}\t\n\\end{align*}","id":"37-1","font":{"size":10,"family":"Arial","color":"#000000"},"type":"align*","ts":1599397187387,"cs":"9INbaG5bZ2ti3e6MkinrpA==","size":{"width":192,"height":132}}

Given {"id":"39-1","font":{"size":10,"color":"#000000","family":"Arial"},"code":"$y=e^{a\\cos^{-1}x}$","type":"$","ts":1599396649181,"cs":"OUFzHMgCUe2IBZuKaO2bqg==","size":{"width":78,"height":17}}

{"id":"41","code":"\\begin{align*}\n{\\diff{y}{x}}&={\\frac{-ay}{{\\sqrt[]{1-x^{2}}}}}\\\\\n{\\left(\\diff{y}{x}\\right)^{2}}&={\\left(\\frac{-ay}{{\\sqrt[]{1-x^{2}}}}\\right)^{2}}\\\\\n{\\left(\\diff{y}{x}\\right)^{2}}&={\\frac{a^{2}y^{2}}{\\left(1-x^{2}\\right)}}\\\\\n\\end{align*}","type":"align*","font":{"family":"Arial","size":10,"color":"#000000"},"ts":1599397831661,"cs":"T7ScZRvKzQGcDe2N+wKybw==","size":{"width":168,"height":130}}

{"font":{"family":"Arial","color":"#000000","size":10},"code":"$\\left(1-x^{2}\\right)\\left(\\diff{y}{x}\\right)^{2}=a^{2}y^{2}$","id":"42","type":"$","ts":1599397879912,"cs":"YnG7VQUZSi5dCgHnT1jfDA==","size":{"width":152,"height":32}}

Again differentiate w.r.t. x:-

{"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\left(1-x^{2}\\right)\\diff{\\left(\\diff{y}{x}\\right)^{2}}{x}+\\left(\\diff{y}{x}\\right)^{2}\\diff{\\left(1-x^{2}\\right)}{x}}&={\\diff{\\left(a^{2}y^{2}\\right)}{x}}\\\\\n{2\\left(1-x^{2}\\right)\\diff{y}{x}\\frac{d^{2}y}{dx^{2}}-2x\\left(\\diff{y}{x}\\right)^{2}}&={2a^{2}y\\diff{y}{x}}\\\\\n{\\left(1-x^{2}\\right)\\diff{y}{x}\\frac{d^{2}y}{dx^{2}}-x\\left(\\diff{y}{x}\\right)^{2}}&={a^{2}y\\diff{y}{x}}\\\\\n{\\left(1-x^{2}\\right)\\frac{d^{2}y}{dx^{2}}-x\\diff{y}{x}}&={a^{2}y}\\\\\n{\\left(1-x^{2}\\right)\\frac{d^{2}y}{dx^{2}}-x\\diff{y}{x}-a^{2}y}&={0}\t\n\\end{align*}","type":"align*","id":"43","ts":1599398169316,"cs":"9pVwSYCWlKkn7ecRkSzlng==","size":{"width":344,"height":225}}

{"font":{"size":10,"color":"#000000","family":"Arial"},"code":"\\begin{align*}\n{\\left(1-x^{2}\\right)\\diff{\\left(\\diff{y}{x}\\right)^{2}}{x}+\\left(\\diff{y}{x}\\right)^{2}\\diff{\\left(1-x^{2}\\right)}{x}}&={\\diff{\\left(a^{2}y^{2}\\right)}{x}}\\\\\n{2\\left(1-x^{2}\\right)\\diff{y}{x}\\frac{d^{2}y}{dx^{2}}-2x\\left(\\diff{y}{x}\\right)^{2}}&={2a^{2}y\\diff{y}{x}}\\\\\n{\\left(1-x^{2}\\right)\\diff{y}{x}\\frac{d^{2}y}{dx^{2}}-x\\left(\\diff{y}{x}\\right)^{2}}&={a^{2}y\\diff{y}{x}}\\\\\n{\\left(1-x^{2}\\right)\\frac{d^{2}y}{dx^{2}}-x\\diff{y}{x}}&={a^{2}y}\\\\\n{\\left(1-x^{2}\\right)\\frac{d^{2}y}{dx^{2}}-x\\diff{y}{x}-a^{2}y}&={0}\t\n\\end{align*}","type":"align*","id":"43","ts":1599398169316,"cs":"9pVwSYCWlKkn7ecRkSzlng==","size":{"width":344,"height":225}}

{"type":"align*","font":{"color":"#000000","size":10,"family":"Arial"},"code":"\\begin{align*}\n{\\left(1-x^{2}\\right)\\frac{d^{2}y}{dx^{2}}-x\\diff{y}{x}}&={a^{2}y}\\\\\n{\\left(1-x^{2}\\right)\\frac{d^{2}y}{dx^{2}}-x\\diff{y}{x}-a^{2}y}&={0}\t\n\\end{align*}","id":"44","ts":1599407661139,"cs":"AionlaeEPORoB3Oa4xYMIA==","size":{"width":226,"height":76}}

(Hence Proved)


Download PDF of Miscellaneous Exercise

See Also:-

Notes of Continuity & Differentiability

Exercise 5.1

Exercise 5.2

Exercise 5.3

Exercise 5.4

Exercise 5.5

Exercise 5.6

Exercise 5.7

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