NCERT Solutions Class-12-Chapter-5-Differentiability and Continuity
Excercise-5.7
Note:
Second Order Derivative: If \( y = f(x) \), then \( \frac{dy}{dx} \) is the first derivative. The derivative of \( \frac{dy}{dx} \) with respect to \( x \) is called the second order derivative.
Notation: \( \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d^2y}{dx^2} \) or \( f''(x) \) or \( y'' \) or \( y_2 \).
Q1
Find the second order derivative of the function: \( x^2 + 3x + 2 \)
Solution:
Let \( y = x^2 + 3x + 2 \).
Differentiating w.r.t. \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(3x) + \frac{d}{dx}(2) = 2x + 3 \).
Differentiating again w.r.t. \( x \):
\( \frac{d^2y}{dx^2} = \frac{d}{dx}(2x + 3) = 2(1) + 0 = 2 \).
Q2
Find the second order derivative of the function: \( x^{20} \)
Solution:
Let \( y = x^{20} \).
\( \frac{dy}{dx} = 20x^{19} \).
\( \frac{d^2y}{dx^2} = \frac{d}{dx}(20x^{19}) = 20(19x^{18}) = 380x^{18} \).
Q3
Find the second order derivative of the function: \( x \cdot \cos x \)
Solution:
Let \( y = x \cos x \).
Using Product Rule for the first derivative:
\[ \frac{dy}{dx} = x \frac{d}{dx}(\cos x) + \cos x \frac{d}{dx}(x) = x(-\sin x) + \cos x (1) = -x \sin x + \cos x \]
Differentiating again:
\[ \begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx}(-x \sin x) + \frac{d}{dx}(\cos x) \\ &= -[x \cos x + \sin x(1)] - \sin x \\ &= -x \cos x - \sin x - \sin x \\ &= -(x \cos x + 2 \sin x) \end{aligned} \]
Q4
Find the second order derivative of the function: \( \log x \)
Solution:
Let \( y = \log x \).
\( \frac{dy}{dx} = \frac{1}{x} = x^{-1} \).
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2} = -\frac{1}{x^2} \]
Q5
Find the second order derivative of the function: \( x^3 \log x \)
Solution:
Let \( y = x^3 \log x \).
Using Product Rule:
\[ \frac{dy}{dx} = x^3 \cdot \frac{1}{x} + \log x \cdot (3x^2) = x^2 + 3x^2 \log x = x^2(1 + 3\log x) \]
Differentiating again:
\[ \begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx}(x^2) + \frac{d}{dx}(3x^2 \log x) \\ &= 2x + 3 \left[ x^2 \cdot \frac{1}{x} + \log x \cdot 2x \right] \\ &= 2x + 3[x + 2x \log x] \\ &= 2x + 3x + 6x \log x \\ &= 5x + 6x \log x = x(5 + 6 \log x) \end{aligned} \]
Q6
Find the second order derivative of the function: \( e^x \sin 5x \)
Solution:
Let \( y = e^x \sin 5x \).
\[ \begin{aligned} \frac{dy}{dx} &= e^x \frac{d}{dx}(\sin 5x) + \sin 5x \frac{d}{dx}(e^x) \\ &= e^x (5 \cos 5x) + e^x \sin 5x \\ &= e^x (5 \cos 5x + \sin 5x) \end{aligned} \]
Differentiating again:
\[ \begin{aligned} \frac{d^2y}{dx^2} &= e^x \frac{d}{dx}(5 \cos 5x + \sin 5x) + (5 \cos 5x + \sin 5x) \frac{d}{dx}(e^x) \\ &= e^x [5(-\sin 5x \cdot 5) + \cos 5x \cdot 5] + e^x (5 \cos 5x + \sin 5x) \\ &= e^x [-25 \sin 5x + 5 \cos 5x + 5 \cos 5x + \sin 5x] \\ &= e^x [10 \cos 5x - 24 \sin 5x] \\ &= 2e^x (5 \cos 5x - 12 \sin 5x) \end{aligned} \]
Q7
Find the second order derivative of the function: \( e^{6x} \cos 3x \)
Solution:
Let \( y = e^{6x} \cos 3x \).
\[ \begin{aligned} \frac{dy}{dx} &= e^{6x}(-3\sin 3x) + \cos 3x (6e^{6x}) \\ &= 3e^{6x}(2\cos 3x - \sin 3x) \end{aligned} \]
Differentiating again:
\[ \begin{aligned} \frac{d^2y}{dx^2} &= 3 \left[ e^{6x} \frac{d}{dx}(2\cos 3x - \sin 3x) + (2\cos 3x - \sin 3x) \frac{d}{dx}(e^{6x}) \right] \\ &= 3 \left[ e^{6x}(-6\sin 3x - 3\cos 3x) + (2\cos 3x - \sin 3x)(6e^{6x}) \right] \\ &= 3e^{6x} \left[ -6\sin 3x - 3\cos 3x + 12\cos 3x - 6\sin 3x \right] \\ &= 3e^{6x} [ 9\cos 3x - 12\sin 3x ] \\ &= 9e^{6x} (3\cos 3x - 4\sin 3x) \end{aligned} \]
Q8
Find the second order derivative of the function: \( \tan^{-1} x \)
Solution:
Let \( y = \tan^{-1} x \).
\( \frac{dy}{dx} = \frac{1}{1+x^2} = (1+x^2)^{-1} \).
\[ \begin{aligned} \frac{d^2y}{dx^2} &= -1(1+x^2)^{-2} \cdot \frac{d}{dx}(1+x^2) \\ &= \frac{-1}{(1+x^2)^2} \cdot (2x) \\ &= \frac{-2x}{(1+x^2)^2} \end{aligned} \]
Q9
Find the second order derivative of the function: \( \log(\log x) \)
Solution:
Let \( y = \log(\log x) \).
\( \frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) = \frac{1}{\log x} \cdot \frac{1}{x} = (x \log x)^{-1} \).
\[ \begin{aligned} \frac{d^2y}{dx^2} &= -1(x \log x)^{-2} \cdot \frac{d}{dx}(x \log x) \\ &= \frac{-1}{(x \log x)^2} \cdot \left( x \cdot \frac{1}{x} + \log x \cdot 1 \right) \\ &= -\frac{1 + \log x}{(x \log x)^2} \end{aligned} \]
Q10
Find the second order derivative of the function: \( \sin(\log x) \)
Solution:
Let \( y = \sin(\log x) \).
\( \frac{dy}{dx} = \cos(\log x) \cdot \frac{1}{x} = \frac{\cos(\log x)}{x} \).
Using Quotient Rule:
\[ \begin{aligned} \frac{d^2y}{dx^2} &= \frac{x \frac{d}{dx}[\cos(\log x)] - \cos(\log x) \frac{d}{dx}[x]}{x^2} \\ &= \frac{x [-\sin(\log x) \cdot \frac{1}{x}] - \cos(\log x)}{x^2} \\ &= \frac{-\sin(\log x) - \cos(\log x)}{x^2} \end{aligned} \]
Q11
If \( y = 5 \cos x - 3 \sin x \), prove that \( \frac{d^2y}{dx^2} + y = 0 \).
Solution:
\( y = 5 \cos x - 3 \sin x \).
\( \frac{dy}{dx} = 5(-\sin x) - 3(\cos x) = -5 \sin x - 3 \cos x \).
\[ \begin{aligned} \frac{d^2y}{dx^2} &= -5(\cos x) - 3(-\sin x) \\ &= -5 \cos x + 3 \sin x \\ &= -(5 \cos x - 3 \sin x) \\ &= -y \end{aligned} \]
Therefore, \( \frac{d^2y}{dx^2} + y = 0 \).
Q12
If \( y = \cos^{-1} x \), find \( \frac{d^2y}{dx^2} \) in terms of \( y \) alone.
Solution:
Given \( y = \cos^{-1} x \implies x = \cos y \).
Differentiating w.r.t. \( x \):
\( 1 = -\sin y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = -\frac{1}{\sin y} = -\text{cosec } y \).
Differentiating again w.r.t. \( x \):
\[ \begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx}(-\text{cosec } y) \\ &= \frac{d}{dy}(-\text{cosec } y) \cdot \frac{dy}{dx} \quad (\text{Chain Rule}) \\ &= -(-\text{cosec } y \cot y) \cdot (-\text{cosec } y) \\ &= -\cot y \cdot \text{cosec}^2 y \end{aligned} \]
Q13
If \( y = 3 \cos(\log x) + 4 \sin(\log x) \), show that \( x^2 y_2 + x y_1 + y = 0 \).
Solution:
\( y = 3 \cos(\log x) + 4 \sin(\log x) \).
\( y_1 = -3 \sin(\log x) \cdot \frac{1}{x} + 4 \cos(\log x) \cdot \frac{1}{x} \).
Multiplying by \( x \):
\( x y_1 = -3 \sin(\log x) + 4 \cos(\log x) \).
Differentiating w.r.t. \( x \) again:
\( x y_2 + y_1(1) = -3 \cos(\log x) \cdot \frac{1}{x} - 4 \sin(\log x) \cdot \frac{1}{x} \).
Multiplying by \( x \) again:
\[ \begin{aligned} x^2 y_2 + x y_1 &= -3 \cos(\log x) - 4 \sin(\log x) \\ x^2 y_2 + x y_1 &= -[3 \cos(\log x) + 4 \sin(\log x)] \\ x^2 y_2 + x y_1 &= -y \end{aligned} \]
Therefore, \( x^2 y_2 + x y_1 + y = 0 \).
Q14
If \( y = A e^{mx} + B e^{nx} \), show that \( \frac{d^2y}{dx^2} - (m+n)\frac{dy}{dx} + mny = 0 \).
Solution:
\( y = A e^{mx} + B e^{nx} \).
\( \frac{dy}{dx} = m A e^{mx} + n B e^{nx} \).
\( \frac{d^2y}{dx^2} = m^2 A e^{mx} + n^2 B e^{nx} \).
Substitute into the LHS:
\[ \begin{aligned} LHS &= (m^2 A e^{mx} + n^2 B e^{nx}) - (m+n)(m A e^{mx} + n B e^{nx}) + mn(A e^{mx} + B e^{nx}) \\ &= m^2 A e^{mx} + n^2 B e^{nx} - (m^2 A e^{mx} + mn B e^{nx} + mn A e^{mx} + n^2 B e^{nx}) + mn A e^{mx} + mn B e^{nx} \\ &= m^2 A e^{mx} + n^2 B e^{nx} - m^2 A e^{mx} - mn B e^{nx} - mn A e^{mx} - n^2 B e^{nx} + mn A e^{mx} + mn B e^{nx} \\ &= 0 = RHS \end{aligned} \]
Q15
If \( y = 500e^{7x} + 600e^{-7x} \), show that \( \frac{d^2y}{dx^2} = 49y \).
Solution:
\( y = 500e^{7x} + 600e^{-7x} \).
\( \frac{dy}{dx} = 500(7)e^{7x} + 600(-7)e^{-7x} = 3500e^{7x} - 4200e^{-7x} \).
\[ \begin{aligned} \frac{d^2y}{dx^2} &= 3500(7)e^{7x} - 4200(-7)e^{-7x} \\ &= 49(500)e^{7x} + 49(600)e^{-7x} \\ &= 49 [500e^{7x} + 600e^{-7x}] \\ &= 49y \end{aligned} \]
Q16
If \( e^y(x+1) = 1 \), show that \( \frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2 \).
Solution:
\( e^y = \frac{1}{x+1} \).
Taking log on both sides: \( y = \log(1) - \log(x+1) = -\log(x+1) \).
\( \frac{dy}{dx} = -\frac{1}{x+1} = -(x+1)^{-1} \).
\( \frac{d^2y}{dx^2} = -(-1)(x+1)^{-2} = \frac{1}{(x+1)^2} \).
Also, \( \left(\frac{dy}{dx}\right)^2 = \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2} \).
Thus, \( \frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2 \).
Q17
If \( y = (\tan^{-1} x)^2 \), show that \( (x^2+1)^2 y_2 + 2x(x^2+1) y_1 = 2 \).
Solution:
\( y = (\tan^{-1} x)^2 \).
\( y_1 = 2(\tan^{-1} x) \cdot \frac{1}{1+x^2} \).
\( (1+x^2) y_1 = 2 \tan^{-1} x \).
Differentiating w.r.t. \( x \) using Product Rule:
\( (1+x^2) y_2 + y_1 (2x) = 2 \cdot \frac{1}{1+x^2} \).
Multiplying both sides by \( (1+x^2) \):
\( (1+x^2)^2 y_2 + 2x(1+x^2) y_1 = 2 \).