#### need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 65

Hint: you must know the rules of derivative of exponential functions.

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

Prove   $\frac{d y}{d x}=1-y^{2}$

Solution:

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

Differentiate with respect to x,use quotient rule

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

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

$\frac{d y}{d x}=\left\{\frac{\left(e^{x}+e^{-x}\right)\left[\left(e^{x}-e^{-x}(-1)\right)\right]-\left(e^{x}-e^{-x}\right)\left[\left(e^{x}+e^{-x}(-1)\right)\right]}{\left(e^{x}+e^{-x}\right)^{2}}\right\}$

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

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

$\frac{d y}{d x}=1-\frac{\left(e^{x}-e^{-x}\right)^{2}}{\left(e^{x}+e^{-x}\right)^{2}}$                        $\left[y=\frac{\left(e^{x}-e^{-x}\right)}{e^{x}+e^{-x}}\right]$

$\frac{d y}{d x}=1-y^{2}$

∴ Proved