Question

If $y=\frac{e^x+e^{-x}}{e^x-e^x}$, then $\frac{dy}{dx}$ is equal to

• A. ${sech}^{2}x$
• B. ${cosech}^{2}x$
• C. $-{sech}^{2}x$
• D. $-{cosech}^{2}x$

Answer 1

The correct option is D

$-{cosech}^{2}x$

Finding the derivative:

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

Divide numerator & denominator by $2.$

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

Now we know that, $\cos \mathrm{hx}=\frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2},\sin \mathrm{hx}=\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}}{2}$.

$$\begin{gathered} \Rightarrow \mathrm{y}=\frac{\cos \mathrm{hx}}{\sin \mathrm{hx}} \\ \Rightarrow \mathrm{y}=\cot \mathrm{hx} \end{gathered}$$

Now differentiate both sides with respect to $x$ we have:

$\frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{cosec}{\mathrm{h}}^2\mathrm{x}[\frac{d\mathrm{y}}{d\mathrm{x}}=-\mathrm{cosec}{\mathrm{h}}^2\mathrm{x}]$

Hence, option (D) is correct.