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

$y = {({\tan ^{ - 1}}x)^2}$ $Diff.w.r.t.x,$ ${y_1} = 2({\tan ^{ - 1}}x)\frac{d}{{dx}}({\tan ^{ - 1}}x)$ ${y_1} = 2({\tan ^{ - 1}}x)\frac{1}{{1 + {x^2}}}$ $\left( {1 + {x^2}} \right){y_1} = 2({\tan ^{ - 1}}x)$ $Diff.w.r.t.x,again$ $\left( {1 + {x^2}} \right){y_2} + 2x{y_1} = 2\frac{1}{{1 + {x^2}}}$ ${\left( {1 + {x^2}} \right)^2}{y_2} + 2x(1 + {x^2}){y_1} = 2$