Solve this

Solution:

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

On differentiating $y$ with respect to $x$, we get

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

Recall that $\left(\frac{\mathrm{u}}{\mathrm{v}}\right)^{\prime}=\frac{\mathrm{vu}^{\prime}-\mathrm{uv}^{\prime}}{\mathrm{v}^{2}}$ (quotient rule)

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

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

We know $\frac{d}{d x}\left(e^{x}\right)=e^{x}$

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

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

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

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

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

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=1-\mathrm{y}^{2}$

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