Exercise 5.7 Class 12 Maths Continuity and Differentiability

NCERT Solutions Class 12 Maths Chapter 5 Ex 5.7 Detailed | LearnCBSEHub

💡 Key Concept: Second Derivative

If $y = f(x)$, then:

First Derivative: $\frac{dy}{dx} = f'(x) = y_1 = y’$

Second Derivative: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = f”(x) = y_2 = y”$

Questions 1 – 5

Find the second order derivatives of the functions.

Q1: $y = x^2 + 3x + 2$
$\frac{dy}{dx} = 2x + 3 \implies \frac{d^2y}{dx^2} = 2$

Q2: $y = x^{20}$
$\frac{dy}{dx} = 20x^{19} \implies \frac{d^2y}{dx^2} = 380x^{18}$

Q3: $y = x \cos x$
$y’ = \cos x – x \sin x$
$y” = -\sin x – (1 \cdot \sin x + x \cos x) = -2\sin x – x\cos x$

Q4: $y = \log x$
$y’ = \frac{1}{x} \implies y” = -\frac{1}{x^2}$

Q5: $y = x^3 \log x$
$y’ = 3x^2 \log x + x^3 \cdot \frac{1}{x} = 3x^2 \log x + x^2 = x^2(3 \log x + 1)$
$y” = 2x(3 \log x + 1) + x^2 \cdot \frac{3}{x} = 6x \log x + 2x + 3x = x(5 + 6 \log x)$

Question 06

Find $y”$ for $y = e^x \sin 5x$

Step 1: First Derivative

$$\frac{dy}{dx} = e^x \sin 5x + e^x (5 \cos 5x) = e^x(\sin 5x + 5 \cos 5x)$$

Step 2: Second Derivative

$$\frac{d^2y}{dx^2} = e^x(\sin 5x + 5 \cos 5x) + e^x(5 \cos 5x – 25 \sin 5x)$$ $$= e^x(\sin 5x + 10 \cos 5x – 25 \sin 5x)$$ $$= 2e^x(5 \cos 5x – 12 \sin 5x)$$

Answer: $2e^x(5 \cos 5x – 12 \sin 5x)$

Question 07

Find $y”$ for $y = e^{6x} \cos 3x$

Step 1: First Derivative

$$y’ = 6e^{6x}\cos 3x – 3e^{6x}\sin 3x = 3e^{6x}(2\cos 3x – \sin 3x)$$

Step 2: Second Derivative

$$y” = 3[6e^{6x}(2\cos 3x – \sin 3x) + e^{6x}(-6\sin 3x – 3\cos 3x)]$$ $$= 3e^{6x}[12\cos 3x – 6\sin 3x – 6\sin 3x – 3\cos 3x]$$ $$= 3e^{6x}[9\cos 3x – 12\sin 3x] = 9e^{6x}(3\cos 3x – 4\sin 3x)$$

Answer: $9e^{6x}(3\cos 3x – 4\sin 3x)$

Questions 08 – 10

Find second order derivatives.

Q8: $y = \tan^{-1} x$
$y’ = \frac{1}{1+x^2} \implies y” = \frac{0 – 1(2x)}{(1+x^2)^2} = \frac{-2x}{(1+x^2)^2}$

Q9: $y = \log(\log x)$
$y’ = \frac{1}{\log x} \cdot \frac{1}{x} = (x \log x)^{-1}$
$y” = -1(x \log x)^{-2} \cdot (\log x + 1) = -\frac{1 + \log x}{(x \log x)^2}$

Q10: $y = \sin(\log x)$
$y’ = \frac{\cos(\log x)}{x}$
$y” = \frac{x(-\sin(\log x) \cdot \frac{1}{x}) – \cos(\log x)}{x^2} = -\frac{\sin(\log x) + \cos(\log x)}{x^2}$

Question 11

If $y = 5 \cos x – 3 \sin x$, prove that $\frac{d^2y}{dx^2} + y = 0$.

Proof

$$y’ = -5 \sin x – 3 \cos x$$ $$y” = -5 \cos x + 3 \sin x$$ $$y” = -(5 \cos x – 3 \sin x)$$ $$y” = -y \implies y” + y = 0$$

Hence Proved.

Question 12

If $y = \cos^{-1} x$, find $\frac{d^2y}{dx^2}$ in terms of $y$ alone.

Step 1: Express x in terms of y

$y = \cos^{-1} x \implies x = \cos y$

Step 2: Differentiate w.r.t x

$$\frac{dx}{dy} = -\sin y \implies \frac{dy}{dx} = -\frac{1}{\sin y} = -\csc y$$ $$\frac{d^2y}{dx^2} = \frac{d}{dx}(-\csc y) = \frac{d}{dy}(-\csc y) \cdot \frac{dy}{dx}$$ $$= (\csc y \cot y) \cdot (-\csc y)$$ $$= -\cot y \csc^2 y$$

Answer: $-\cot y \csc^2 y$

Question 13

If $y = 3 \cos(\log x) + 4 \sin(\log x)$, prove $x^2 y_2 + x y_1 + y = 0$.

Proof

$$y_1 = \frac{-3 \sin(\log x)}{x} + \frac{4 \cos(\log x)}{x}$$ $$x y_1 = -3 \sin(\log x) + 4 \cos(\log x)$$ $$\text{Differentiating w.r.t x: } x y_2 + y_1 = \frac{-3 \cos(\log x)}{x} – \frac{4 \sin(\log x)}{x}$$ $$x^2 y_2 + x y_1 = -(3 \cos(\log x) + 4 \sin(\log x))$$ $$x^2 y_2 + x y_1 = -y \implies x^2 y_2 + x y_1 + y = 0$$

Hence Proved.

Question 14

If $y = Ae^{mx} + Be^{nx}$, prove $y” – (m+n)y’ + mny = 0$.

Proof

$$y’ = mAe^{mx} + nBe^{nx}$$ $$y” = m^2Ae^{mx} + n^2Be^{nx}$$ $$LHS = (m^2Ae^{mx} + n^2Be^{nx}) – (m+n)(mAe^{mx} + nBe^{nx}) + mn(Ae^{mx} + Be^{nx})$$ $$\text{Expanding terms cancels everything out to 0.}$$

Hence Proved.

Question 15

If $y = 500e^{7x} + 600e^{-7x}$, show that $y” = 49y$.

Proof

$$y’ = 500(7)e^{7x} + 600(-7)e^{-7x}$$ $$y” = 500(49)e^{7x} + 600(49)e^{-7x}$$ $$y” = 49(500e^{7x} + 600e^{-7x}) = 49y$$

Hence Proved.

Question 16

If $e^y(x+1) = 1$, show that $\frac{d^2y}{dx^2} = (\frac{dy}{dx})^2$.

Step 1: Simplify and Differentiate

$e^y = \frac{1}{x+1} \implies y = \log(\frac{1}{x+1}) = -\log(x+1)$

$$\frac{dy}{dx} = -\frac{1}{x+1}$$ $$\frac{d^2y}{dx^2} = -(-1)(x+1)^{-2} = \frac{1}{(x+1)^2}$$ $$\left(\frac{dy}{dx}\right)^2 = \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2}$$

Hence Proved: LHS = RHS

Question 17

If $y = (\tan^{-1} x)^2$, show $(x^2+1)^2 y_2 + 2x(x^2+1) y_1 = 2$.

Proof

$$y_1 = 2 \tan^{-1} x \cdot \frac{1}{1+x^2} \implies (1+x^2)y_1 = 2 \tan^{-1} x$$ $$\text{Differentiating again: } (1+x^2)y_2 + y_1(2x) = 2 \cdot \frac{1}{1+x^2}$$ $$\text{Multiplying by } (1+x^2): (1+x^2)^2 y_2 + 2x(1+x^2)y_1 = 2$$

Hence Proved.