CBSE Class 12th Exam Date:17 Feb' 26 - 17 Feb' 26
Continuity means a function does not jump or disappear, while Differentiability means the function does not stumble and keeps going without any sharp or awkward turns. Understanding how functions change is not just about finding their slopes, but we can go one step further and find how those slopes change to get a better look at how the functions behave. This is where the second-order derivative plays an important role in calculus, it helps us to determine the curvature of the function. In exercise 5.7 of the chapter Continuity and Differentiability, we will learn about the concept of the second-order derivative, which can tell us about how the first-order derivative, i.e. the rate of change itself, is changing. This article on the NCERT Solutions for Exercise 5.7 Class 12 Maths Chapter 5 - Continuity and Differentiability provides detailed solutions for the problems given in the exercise, so that students can clear their doubts and get a clear understanding of the method and logic behind these solutions. For syllabus, notes, and PDF, refer to this link: NCERT.
Question:1 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
$y=x^2 + 3x+ 2$
Now, differentiation w.r.t. x
$\frac{dy}{dx}= 2x+3$
Now, second order derivative
$\frac{d^2y}{dx^2}= 2$
Therefore, the second order derivative is $\frac{d^2y}{dx^2}= 2$
Question:2 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
$y=x ^{20}$
Now, differentiation w.r.t. x
$\frac{dy}{dx}= 20x^{19}$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= 20.19x^{18}= 380x^{18}$
Therefore, second-order derivative is $\frac{d^2y}{dx^2}= 380x^{18}$
Question:3 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
$y = x \cos x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}= \cos x + x(-\sin x ) = \cos x-x\sin x$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= -\sin x-(\sin x+x\cos x) = -2\sin x - x\sin x$
Therefore, the second-order derivative is $\frac{d^2y}{dx^2}= -2\sin x - x\sin x$
Question:4 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
$y=\log x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{1}{x}$
Now, second order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{x^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2}= \frac{-1}{x^2}$
Question:5 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
$y=x^3\log x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=3x^2.\log x+x^3.\frac{1}{x}= 3x^2.\log x+ x^2$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}= 6x.\log x+3x^2.\frac{1}{x}+2x=6x.\log x+3x+2x = x(6.\log x+5)$
Therefore, the second-order derivative is $\frac{d^2y}{dx^2} = x(6.\log x+5)$
Question:6 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
$y= e^x\sin 5x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=e^x.\sin 5x +e^x.5\cos 5x = e^x(\sin5x+5\cos5x)$
Now, second order derivative is
$\frac{d^2y}{dx^2}= e^x(\sin5x+5\cos5x)+e^x(5\cos5x+5.(-5\sin5x))$
$= e^x(\sin5x+5\cos5x)+e^x(5\cos5x-25\sin5x)=e^x(10\cos5x-24\sin5x)$
$=2e^x(5\cos5x-12\sin5x)$
Therefore, second order derivative is $\frac{dy}{dx}=2e^x(5\cos5x-12\sin5x)$
Question:7 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
$y= e^{6x}\cos 3x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=6e^{6x}.\cos 3x +e^{6x}.(-3\sin 3x)= e^{6x}(6\cos 3x-3\sin 3x)$
Now, second order derivative is
$\frac{d^2y}{dx^2}= 6e^{6x}(6\cos3x-3\sin3x)+e^{6x}(6.(-3\sin3x)-3.3\cos3x)$
$= 6e^{6x}(6\cos3x-3\sin3x)-e^{6x}(18\sin3x+9\cos3x)$
$e^{6x}(27\cos3x-36\sin3x) = 9e^{6x}(3\cos3x-4\sin3x)$
Therefore, second order derivative is $\frac{dy}{dx} = 9e^{6x}(3\cos3x-4\sin3x)$
Question:8 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
$y = \tan^{-1}x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\tan^{-1}x)}{dx}=\frac{1}{1+x^2}$
Now, second order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{(1+x^2)^2}.2x = \frac{-2x}{(1+x^2)^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-2x}{(1+x^2)^2}$
Question:9 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
$y = \log(\log x)$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\log(\log x))}{dx}=\frac{1}{\log x}.\frac{1}{x}= \frac{1}{x\log x}$
Now, second order derivative is
$\frac{d^2y}{dx^2}= \frac{-1}{(x\log x)^2}.(1.\log x+x.\frac{1}{x}) = \frac{-(\log x+1)}{(x\log x)^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-(\log x+1)}{(x\log x)^2}$
Question:10 Find the second order derivatives of the functions given in Exercises 1 to 10.
Answer:
Given function is
$y = \sin(\log x)$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(\sin(\log x))}{dx}=\cos (\log x).\frac{1}{x}= \frac{\cos (\log x)}{x}$
Now, second order derivative is
Using Quotient rule
$\frac{d^2y}{dx^2}=\frac{-\sin(\log x)\frac{1}{x}.x-\cos(\log x).1}{x^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}$
Therefore, second order derivative is $\frac{d^2y}{dx^2} = \frac{-(\sin (\log x)+\cos(\log x))}{x^2}$
Question:11 If $y = 5 \cos x - 3 \sin x$ prove that $\frac{d^2y}{dx^2}+y = 0$
Answer:
Given function is
$y = 5 \cos x - 3 \sin x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(5\cos x-3\sin x)}{dx}=-5\sin x-3\cos x$
Now, the second-order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(-5\sin x-3\cos x)}{dx^2}=-5\cos x+3\sin x$
Now,
$\frac{d^2y}{dx^2}+y=-5\cos x+3\sin x+5\cos x-3\sin x = 0$
Hence proved
Question:12 If $y = \cos ^{-1} x$ Find $\frac{d ^2 y }{dx^2 }$ in terms of y alone.
Answer:
Given function is
$y = \cos ^{-1} x$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d( \cos ^{-1} x)}{dx}=\frac{-1}{\sqrt{1-x^2}}$
Now, second order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{\sqrt{1-x^2}})}{dx^2}=\frac{-(-1)}{(\sqrt{1-x^2})^2}.(-2x) = \frac{-2x}{1-x^2}$ -(i)
Now, we want $\frac{d^2y}{dx^2}$ in terms of y
$y = \cos ^{-1} x$
$x = \cos y$
Now, put the value of x in (i)
$\frac{d^2y}{dx^2} = \frac{-2\cos y}{1 - \cos^2 y} = \frac{-2\cos y}{\sin^2 y} = -2\cot y \, \operatorname{cosec} y$
$\frac{d^2y}{dx^2} = \frac{-2\cos y}{1 - \cos^2 y} = \frac{-2\cos y}{\sin^2 y} = -2\cot y\, \mathrm{cosec}\, y$
$\left(\because\ 1 - \cos^2 x = \sin^2 x,\ \frac{\cos x}{\sin x} = \cot x,\ \text{and}\ \frac{1}{\sin x} = \mathrm{cosec}\, x \right)$
Therefore, answer is $\frac{d^2y}{dx^2} = -2 \cot y \, \mathrm{cosec}\, y$
Question:13 If $y = 3 \cos (\log x) + 4 \sin (\log x)$, show that $x^2 y_2 + xy_1 + y = 0$
Answer:
Given function is
$y = 3 \cos (\log x) + 4 \sin (\log x)$
Now, differentiation w.r.t. x
$y_1=\frac{dy}{dx}=\frac{d( 3 \cos (\log x) + 4 \sin (\log x))}{dx}=-3\sin(\log x).\frac{1}{x}+4\cos (\log x).\frac{1}{x}$
$=\frac{4\cos (\log x)-3\sin(\log x)}{x}$ -(i)
Now, second order derivative is
By using the Quotient rule
$y_2 = \frac{d^2y}{dx^2} = \frac{d^2\left(\frac{4\cos(\log x) - 3\sin(\log x)}{x}\right)}{dx^2}$
$= \frac{\left(-4\sin(\log x) \cdot \frac{1}{x} - 3\cos(\log x) \cdot \frac{1}{x}\right) \cdot x - 1 \cdot \left(4\cos(\log x) - 3\sin(\log x)\right)}{x^2}$
$= \frac{-\sin(\log x) + 7\cos(\log x)}{x^2} \ \text{-(ii)}$
Now, from equation (i) and (ii) we will get $y_1 \ and \ y_2$
Now, we need to show
$x^2 y_2 + xy_1 + y = 0$
Put the value of $y_1 \ and \ y_2$ from equation (i) and (ii)
$x^2\left( \frac{-\sin(\log x) + 7\cos(\log x)}{x^2} \right) + x\left( \frac{4\cos(\log x) - 3\sin(\log x)}{x} \right) + 3\cos(\log x) + 4\sin(\log x)$
$-\sin(\log x) - 7\cos(\log x) + 4\cos(\log x) - 3\sin(\log x) + 3\cos(\log x) + 4\sin(\log x)$
$=0$
Hence proved
Question:14 If $y = A e ^{mx} + Be ^{nx}$ , show that $\frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0$
Answer:
Given function is
$y = A e ^{mx} + Be ^{nx}$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(A e ^{mx} + Be ^{nx})}{dx}=mAe^{mx}+nBe^{nx}$ -(i)
Now, second order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(mAe^{mx}+nBe^{nx})}{dx^2}= m^2Ae^{mx}+n^2Be^{nx}$ -(ii)
Now, we need to show
$\frac{d ^2 y}{dx^2} - (m+n) \frac{dy}{dx} + mny = 0$
Put the value of $\frac{d^2y}{dx^2} \ and \ \frac{dy}{dx}$ from equation (i) and (ii)
$m^2Ae^{mx}+n^2Be^{nx}-(m+n)(mAe^{mx}+nBx^{nx}) +mn(Ae^{mx}+Be^{nx})$
$m^2Ae^{mx}+n^2Be^{nx}-m^2Ae^{mx}-mnBx^{nx}-mnAe^{mx} -n^2Be^{nx}+mnAe^{mx}$$+mnBe^{nx}$
$=0$
Hence proved
Question:15 If $y = 500 e ^{7x} + 600 e ^{- 7x }$ , show that $\frac{d^2 y}{dx ^2} = 49 y$
Answer:
Given function is
$y = 500 e ^{7x} + 600 e ^{- 7x }$
Now, differentiation w.r.t. x
$\frac{dy}{dx}=\frac{d(500 e ^{7x} + 600 e ^{- 7x })}{dx}=7.500e^{7x}-7.600e^{-7x} =3500e^{7x}-4200e^{-7x}$ -(i)
Now, second order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(3500e^{7x}-4200e^{-7x})}{dx^2}$
$= 7.3500e^{7x}-(-7).4200e^{-7x}= 24500e^{7x}+29400e^{-7x}$ -(ii)
Now, we need to show
$\frac{d^2 y}{dx ^2} = 49 y$
Put the value of $\frac{d^2y}{dx^2}$ from equation (ii)
$24500e^{7x}+29400e^{-7x}=49(500e^{7x}+600e^{-7x})$
$= 24500e^{7x}+29400e^{-7x}$
Hence, L.H.S. = R.H.S.
Hence proved
Question:16 If $e ^y (x+1) = 1$ show that $\frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2$
Answer:
Given function is
$e ^y (x+1) = 1$
We can rewrite it as
$e^y = \frac{1}{x+1}$
Now, differentiation w.r.t. x
$\frac{d(e^y)}{dx}=\frac{d(\frac{1}{x+1})}{dx}\\ e^y.\frac{dy}{dx}= \frac{-1}{(x+1)^2}\\ \frac{1}{x+1}.\frac{dy}{dx}= \frac{-1}{(x+1)^2} \ \ \ \ \ \ \ \ \ (\because e^y = \frac{1}{x+1})\\ \frac{dy}{dx}= \frac{-1}{x+1}$ -(i)
Now, second order derivative is
$\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{x+1})}{dx^2}=\frac{-(-1)}{(x+1)^2} = \frac{1}{(x+1)^2}$ -(ii)
Now, we need to show
$\frac{d^2 y }{dx^2 } = (\frac{dy}{dx})^2$
Put value of $\frac{d^2y}{dx^2} \ and \ \frac{dy}{dx}$ from equation (i) and (ii)
$\frac{1}{(x+1)^2}=\left ( \frac{-1}{x+1} \right )^2$
$=\frac{1}{(x+1)^2}$
Hence, L.H.S. = R.H.S.
Hence proved
Question:17 If $y = (\tan^{-1} x)^2$ show that $(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2$
Answer:
Given function is
$y = (\tan^{-1} x)^2$
Now, differentiation w.r.t. x
$y_1=\frac{dy}{dx}=\frac{d((\tan^{-1}x)^2)}{dx}= 2.\tan^{-1}x.\frac{1}{1+x^2}= \frac{2\tan^{-1}x}{1+x^2}$ -(i)
Now, the second-order derivative is
By using the quotient rule
$y_2=\frac{d^2y}{dx^2}=\frac{d^2(\frac{2\tan^{-1}x}{1+x^2})}{dx^2}=\frac{2.\frac{1}{1+x^2}.(1+x^2)-2\tan^{-1}x(2x)}{(1+x^2)^2}=\frac{2-4x\tan^{-1}x}{(1+x^2)^2}$ -(ii)
Now, we need to show
$(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2$
Put the value from equation (i) and (ii)
$(x^2+1)^2.\frac{2-4x\tan^{-1}x}{(1+x^2)^2}+2x(x^2+1).\frac{2\tan^{-1}x}{x^2+1}\\ \Rightarrow 2-4x\tan^{-1}x+4x\tan^{-1}x = 2$
Hence, L.H.S. = R.H.S.
Hence proved
Also Read,
The main topics covered in Chapter 5 of continuity and differentiability, exercises 5.7 are:
Also Read,
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Frequently Asked Questions (FAQs)
y = c
dy/dx = 0
y = c
dy/dx = 0
d(dy/dx)/dx = 0
Given y = x
dy/dx = 1
Given y = x
dy/dx = 1
d(dy/dx)/dx = 0
y = e^x
dy/dx = e^x
d(dy/dx)/dx = e^x
d^(2)y/dx^2 = e^x
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