#### need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 31

Answer: $\frac{-8}{\left(e^{2 x}-e^{-2 x}\right)^{2}}$

Hint: You must know the rules of solving derivative of exponential function.

Given:  $\frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}$

Solution:

Let  $y=\frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}$

Differentiate with respect to x,

$\frac{d y}{d x}=\frac{d}{d x}\left[\frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}\right]$

$\frac{d y}{d x}=\left\{\frac{\left(e^{2 x}-e^{-2 x}\right) \frac{d}{d x}\left(e^{2 x}+e^{-2 x}\right)-\left(e^{2 x}+e^{-2 x}\right) \frac{d}{d x}\left(e^{2 x}-e^{-2 x}\right)}{\left(e^{2 x}+e^{-2 x}\right)^{2}}\right\}$$\cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$

Using quotient rule,

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

$\frac{d y}{d x}=\frac{2\left(e^{4 x}+e^{-4 x}-2 e^{2 x} e^{-2 x}-e^{4 x}-e^{-4 x}-2 e^{2 x} e^{-2 x}\right)}{\left(e^{2 x}+e^{-2 x}\right)^{2}}$

$\frac{d y}{d x}=\frac{-8}{\left(e^{2 x}+e^{-2 x}\right)^{2}}$

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support