12th NCERT Continuity And Differentiability Exercise 5.7 Questions 17
Hi
Find the second order derivative of the function given in Exercise 1 to 10.

Question (1)

x2 + 3x + 2

Solution

The derivative of the derivative of the function is called second order derivative of the function. It is denoted as y'' or \[\frac{{{d^2}y}}{{d{x^2}}}\] or \[{y_2}\] .
\[y = {x^2} + 3x + 2\] \[Diff.w.r.t.x,\] \[\frac{{dy}}{{dx}} = 2x + 3\] \[Diff.w.r.t.x,again\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\] \[ = 2\]

Question (2)

x20

Solution

\[y = {x^{20}}\] \[Diff.w.r.t.x,\] \[\frac{{dy}}{{dx}} = 20\left( {{x^{19}}} \right)\] \[Diff.w.r.t.x,again\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\] \[ = 20\left( {19{x^{18}}} \right)\] \[ = 380{x^{18}}\]

Question (3)

x cos x

Solution

\[y = x\cos x\] \[Diff.w.r.t.x,\] \[\frac{{dy}}{{dx}} = x\frac{d}{{dx}}\cos x + \cos x\frac{d}{{dx}}x\] \[ = x( - \sin x) + \cos x\] \[ = - x\sin x + \cos x\] \[Diff.w.r.t.x,again\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\] \[ = - \left[ {x\frac{d}{{dx}}\sin x + \sin x\frac{d}{{dx}}x} \right] + ( - \sin x)\] \[\begin{array}{l} = - x\cos x - \sin x - \sin x\\ = - x\cos x - 2\sin x\end{array}\]

Question (4)

log x

Solution

\[y = \log x\] \[Diff.w.r.t.x,\] \[\frac{{dy}}{{dx}} = \frac{1}{x}\] \[Diff.w.r.t.x,again\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\] \[ = - \frac{1}{{{x^2}}}\]

Question (5)

x3 log x

Solution

\[y = {x^3}\log x\] \[Diff.w.r.t.x,\] \[\frac{{dy}}{{dx}} = {x^3}\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}{x^3}\] \[ = {x^3}\frac{1}{x} + \log x(3{x^2})\] \[ = {x^2} + 3{x^2}\log x\] \[Diff.w.r.t.x,again\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\] \[ = 2x + 3\left[ {{x^2}\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}{x^2}} \right]\] \[ = 2x + 3\left[ {{x^2}\frac{1}{x} + \log x(2x)} \right]\] \[\begin{array}{l} = 2x + 3x + 6x\log x\\ = 5x + 6x\log x\end{array}\] \[ = x\left( {5 + 6\log x} \right)\]

Question (6)

ex sin5x

Solution

\[y = {e^x}\sin 5x\] \[Diff.w.r.t.x,again\] \[\frac{{dy}}{{dx}} = {e^x}\frac{d}{{dx}}\sin 5x + \sin 5x\frac{d}{{dx}}{e^x}\] \[ = {e^x}\cos 5x(5) + \sin 5x({e^x})\] \[ = 5{e^x}\cos 5x + {e^x}\sin 5x\] \[Diff.w.r.t.x,again\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\] \[ = 5\left[ {{e^x}\frac{d}{{dx}}\cos 5x + \cos 5x\frac{d}{{dx}}{e^x}} \right] + \left[ {{e^x}\frac{d}{{dx}}\sin 5x + \sin 5x\frac{d}{{dx}}{e^x}} \right]\] \[ = 5\left[ {{e^x}( - \sin 5x(5) + \cos 5x{e^x}} \right] + \left[ {{e^x}\cos 5x(5) + \sin 5x{e^x}} \right]\] \[ = - 25\sin 5x{e^x} + 5\cos 5x{e^x} + 5\cos 5x{e^x} + \sin 5x{e^x}\] \[ = - 24{e^x}\sin 5x + 10{e^x}\cos 5x\] \[ = 2{e^x}\left( {5\cos 5x - 12\sin 5x} \right)\]

Question (7)

e6x cos 3x

Solution

\[y = {e^{6x}}\cos 3x\] \[Diff.w.r.t.x,again\] \[\frac{{dy}}{{dx}} = {e^{6x}}\frac{d}{{dx}}\cos 3x + \cos 3x\frac{d}{{dx}}{e^{6x}}\] \[ = {e^{6x}}( - \sin 3x(3) + \cos 3x({e^{6x}})(6)\] \[ = - 3{e^{6x}}\sin 3x + 6{e^{6x}}\cos 3x\] \[ = 3{e^{6x}}\left( {2\cos 3x - \sin 3x} \right)\] \[Diff.w.r.t.x,again\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\] \[ = 3\left[ {{e^{6x}}\frac{d}{{dx}}\left( {2\cos 3x - \sin 3x} \right) + \left( {2\cos 3x - \sin 3x} \right)\frac{d}{{dx}}{e^{6x}}} \right]\] \[ = 3\left[ {{e^{6x}}\left( { - 2\sin 3x(3) - \cos 3x(3)} \right) + \left( {2\cos 3x - \sin 3x} \right){e^{6x}}(6)} \right]\] \[ = 3{e^{6x}}\left[ { - 6\sin 3x - 3\cos 3x + 12\cos 3x - 6\sin 3x} \right]\] \[ = 3{e^{6x}}\left[ {9\cos 3x - 12\sin 3x} \right]\] \[ = 9{e^{6x}}\left( {3\cos 3x - 4\sin 3x} \right)\]

Question (8)

tan-1 x

Solution

\[y = {\tan ^{ - 1}}x\] \[Diff.w.r.t.x,\] \[\frac{{dy}}{{dx}} = \frac{1}{{1 + {x^2}}}\] \[Diff.w.r.t.x,again\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\] \[ = \frac{{ - 1}}{{{{\left( {1 + {x^2}} \right)}^2}}}\frac{d}{{dx}}\left( {1 + {x^2}} \right)\] \[ = \frac{{ - 2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}\]

Question (9)

Log(log x)

Solution

\[y = \log (\log x)\] \[Diff.w.r.t.x,\] \[\frac{{dy}}{{dx}} = \frac{1}{{\log x}}\frac{d}{{dx}}\log x\] \[ = \frac{1}{{\log x}} \times \frac{1}{x}\] \[ = \frac{1}{{x\log x}}\] \[Diff.w.r.t.x,again\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\] \[ = \frac{{ - 1}}{{{{\left( {x\log x} \right)}^2}}}\frac{d}{{dx}}\left( {x\log x} \right)\] \[ = \frac{{ - 1}}{{{{\left( {x\log x} \right)}^2}}}\left[ {x\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}x} \right]\] \[ = \frac{{ - 1}}{{{{\left( {x\log x} \right)}^2}}}\left[ {x \times \frac{1}{x} + \log x} \right]\] \[ = \frac{{ - (1 + \log x)}}{{{{\left( {x\log x} \right)}^2}}}\]

Question (10)

sin(log x)

Solution

\[y = \sin (\log x)\] \[Diff.w.r.t.x,\] \[\frac{{dy}}{{dx}} = \cos (\log x)\frac{d}{{dx}}\log x\] \[ = \frac{{\cos (\log x)}}{x}\] \[Diff.w.r.t.x,again\] \[\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\] \[ = \frac{{x\frac{d}{{dx}}\cos (\log x) - \cos (\log x)\frac{d}{{dx}}x}}{{{x^2}}}\] \[ = \frac{{x( - \sin (\log x))\frac{1}{x} - \cos (\log x)}}{{{x^2}}}\] \[ = \frac{{ - \left[ {\cos (\log x) + \sin (\log x)} \right]}}{{{x^2}}}\]

Question (11)

\[If\;y = 5\cos x - 3\sin x,prove\;that\frac{{{d^2}y}}{{d{x^2}}} + y = 0\]

Solution

\[y = 5\cos x - 3\sin x\] \[Diff.w.r.t.x,\] \[\frac{{dy}}{{dx}} = 5( - \sin x) - 3(\cos x)\] \[\frac{{dy}}{{dx}} = - 5\sin x - 3\cos x\] \[Diff.w.r.t.x,again\] \[\frac{{{d^2}y}}{{d{x^2}}} = - 5\cos x - 3( - \sin x)\] \[\frac{{{d^2}y}}{{d{x^2}}} = - 5\cos x + 3(\sin x)\] \[\frac{{{d^2}y}}{{d{x^2}}} = - \left[ {5\cos x - 3(\sin x)} \right]\] \[\frac{{{d^2}y}}{{d{x^2}}} = - y\] \[\frac{{{d^2}y}}{{d{x^2}}} + y = 0\]

Question (12)

If y = cos-1x, find d2y / dx2 in term of y alone.

Solution

\[y = {\cos ^{ - 1}}x\] \[Diff.w.r.t.x,\] \[\frac{{dy}}{{dx}} = \frac{{ - 1}}{{\sqrt {1 - {x^2}} }}\] \[\frac{{dy}}{{dx}} = - {\left( {1 - {x^2}} \right)^{ - 1/2}}\] \[Diff.w.r.t.x,again\] \[\frac{{{d^2}y}}{{d{x^2}}} = - \left[ { - \frac{1}{2}{{\left( {1 - {x^2}} \right)}^{ - 3/2}} \times - 2x} \right]\] \[\frac{{{d^2}y}}{{d{x^2}}} = - \frac{x}{{{{\left( {1 - {x^2}} \right)}^{3/2}}}}\] \[{\cos ^{ - 1}}x = y \Rightarrow x = \cos y\] Replacing x as cos y we get, \[\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{\cos y}}{{{{\left( {1 - {{\cos }^2}y} \right)}^{3/2}}}}\] \[\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{\cos y}}{{{{\left( {{{\sin }^2}y} \right)}^{3/2}}}}\] \[\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{\cos y}}{{\left( {{{\sin }^3}y} \right)}}\] \[\frac{{{d^2}y}}{{d{x^2}}} = - \cot y\cos e{c^2}y\]

Question (13)

\[If\;y = 3\cos (\log x) + 4\sin (\log x),show\;that\;{x^2}{y_2} + x{y_1} + y = 0\]

Solution

\[y = 3\cos (\log x) + 4\sin (\log x)\] \[Diff.w.r.t.x,\] \[\frac{{dy}}{{dx}} = 3\left[ { - \sin (\log x)\frac{1}{x}} \right] + 4\left[ {\cos (\log x)\frac{1}{x}} \right]\] \[{y_1} = \frac{{ - 3\sin (\log x) + 4\cos (\log x)}}{x}\] \[x{y_1} = - 3\sin (\log x) + 4\cos (\log x)\] \[Diff.w.r.t.x,again\] \[x{y_2} + {y_1} = - 3\cos (\log x)\frac{1}{x} + 4( - \sin (\log x))\frac{1}{x}\] \[{x^2}{y_2} + x{y_1} = - 3\cos (\log x) - 4\sin (\log x)\] \[{x^2}{y_2} + x{y_1} = - \left[ {3\cos (\log x) + 4\sin (\log x)} \right]\] \[{x^2}{y_2} + x{y_1} = - y\] \[{x^2}{y_2} + x{y_1} + y = 0\]

Question (14)

\[If\;y = A{e^{mx}} + B{e^{nx}},show\;that\frac{{{d^2}y}}{{d{x^2}}} - (m + n)\frac{{dy}}{{dx}} + mny = 0\]

Solution

\[y = A{e^{mx}} + B{e^{nx}}\] \[Diff.w.r.t.x\] \[{y_1} = A{e^{mx}}m + B{e^{nx}}n\] \[{y_1} = Am{e^{mx}} + Bn{e^{nx}}\] \[{y_1} = m[A{e^{mx}} + B{e^{nx}}] - Bm{e^{nx}} + Bn{e^{nx}}\] \[{y_1} = my - B{e^{nx}}(m - n)\] \[{y_1} - my = - B(m - n){e^{nx}} - - - (1)\] \[Diff.w.r.t.x,again\] \[{y_2} - m{y_1} = - B(m - n){e^{nx}}n\] \[{y_2} - m{y_1} = n\left[ { - B(m - n){e^{nx}}} \right]\] By replacing the value from equation (1) \[{y_2} - m{y_1} = n({y_1} - my)\] \[{y_2} - m{y_1} = n{y_1} - mny\] \[{y_2} - m{y_1} - n{y_1} + mny = 0\] \[{y_2} - (m + n){y_1} + mny = 0\]

Question (15)

\[If\;y = 500{e^{7x}} + 600{e^{ - 7x}},show\;that\frac{{{d^2}y}}{{d{x^2}}} = 49y\]

Solution

\[y = 500{e^{7x}} + 600{e^{ - 7x}}\] \[Diff.w.r.t.x,\] \[{y_1} = 500{e^{7x}}(7) + 600{e^{ - 7x}}( - 7)\] \[{y_1} = 3500{e^{7x}} - 4200n{e^{ - 7x}}\] \[Diff.w.r.t.x,again\] \[{y_2} = 3500{e^{7x}}(7) - 4200{e^{ - 7x}}( - 7)\] \[{y_2} = 24500{e^{7x}} + 29400{e^{ - 7x}}\] \[{y_2} = 49(500{e^{7x}} + 600{e^{ - 7x}})\] \[{y_2} = 49y\] \[\frac{{{d^2}y}}{{d{x^2}}} = 49y\]

Question (16)

\[If\;{e^y}(x + 1) = 1,show\;that\frac{{{d^2}y}}{{d{x^2}}} = {\left( {\frac{{dy}}{{dx}}} \right)^2}\]

Solution

\[{e^y}(x + 1) = 1\] \[{e^y} = \frac{1}{{x + 1}} - - - (1)\] \[Diff.w.r.t.x,\] \[{e^y}\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{{(x + 1)}^2}}}\] Replacing the value of ey from equation (1) we get, \[\frac{1}{{(x + 1)}}\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{{(x + 1)}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{ - 1}}{{(x + 1)}}\] \[Diff.w.r.t.x,again\] \[\frac{{{d^2}y}}{{d{x^2}}} = - \left( {\frac{{ - 1}}{{{{(x + 1)}^2}}}} \right)\] \[\frac{{{d^2}y}}{{d{x^2}}} = \left( {\frac{1}{{{{(x + 1)}^2}}}} \right)\] \[\frac{{{d^2}y}}{{d{x^2}}} = {\left( {\frac{{ - 1}}{{(x + 1)}}} \right)^2}\] \[\frac{{{d^2}y}}{{d{x^2}}} = {\left( {\frac{{dy}}{{dx}}} \right)^2}\]

Question (17)

\[If\;y = {({\tan ^{ - 1}}x)^2},show\;that{({x^2} + 1)^2}{y_2} + 2x({x^2} + 1){y_1} = 2\]

Solution

\[y = {({\tan ^{ - 1}}x)^2}\] \[Diff.w.r.t.x,\] \[{y_1} = 2({\tan ^{ - 1}}x)\frac{d}{{dx}}({\tan ^{ - 1}}x)\] \[{y_1} = 2({\tan ^{ - 1}}x)\frac{1}{{1 + {x^2}}}\] \[\left( {1 + {x^2}} \right){y_1} = 2({\tan ^{ - 1}}x)\] \[Diff.w.r.t.x,again\] \[\left( {1 + {x^2}} \right){y_2} + 2x{y_1} = 2\frac{1}{{1 + {x^2}}}\] \[{\left( {1 + {x^2}} \right)^2}{y_2} + 2x(1 + {x^2}){y_1} = 2\]
Exercise 5.6 ⇐
⇒ Exercise 5.8