Differentiate the following functions with respect to x :

Solution:

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

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

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

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$

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

However, $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$

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

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

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

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

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

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

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-8}{\left(\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{-2 \mathrm{x}}\right)^{2}}$

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