Question

If $y=\frac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}$, then $\frac{dy}{dx}=?$

• A. $\frac{-8}{{\left(e^{2x}-e^{-2x}\right)}^2}$
• B. $\frac{8}{{\left(e^{2x}-e^{-2x}\right)}^2}$
• C. $\frac{-4}{{\left(e^{2x}-e^{-2x}\right)}^2}$
• D. $\frac{4}{{\left(e^{2x}-e^{-2x}\right)}^2}$

Answer 1

The correct option is A

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

Explanation for the correct option:

Finding the value of $\frac{dy}{dx}$:

Given,

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

Now differentiate with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{\left(2e^{2x}-2e^{-2x}\right)\left(e^{2x}-e^{-2x}\right)-\left(e^{2x}+e^{-2x}\right)\left(2e^{2x}+2e^{-2x}\right)}{{\left(e^{2x}-e^{-2x}\right)}^2}\mathbf{[}\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\frac{\mathbf{u}}{\mathbf{v}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{u}\mathbf{\text{'}}\mathbf{v}\mathbf{}\mathbf{-}\mathbf{}\mathbf{v}\mathbf{\text{'}}\mathbf{u}}{{\mathbf{v}}^{\mathbf{2}}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\frac{\mathbf{d}\mathbf{\left(}{\mathbf{e}}^{\mathbf{x}}\mathbf{\right)}}{\mathbf{d}\mathbf{x}}\mathbf{=}{\mathbf{e}}^{\mathbf{x}}\mathbf{]}\\ & =& \frac{2\left(e^{2x}-e^{-2x}\right)\left(e^{2x}-e^{-2x}\right)-2\left(e^{2x}+e^{-2x}\right)\left(e^{2x}+e^{-2x}\right)}{{\left(e^{2x}-e^{-2x}\right)}^2}\\ & =& \frac{2\left[e^{4x}-e^0-e^0+e^{-4x}-\left(e^{4x}+e^0+e^0+e^{-4x}\right)\right]}{{\left(e^{2x}-e^{-2x}\right)}^2}\\ & =& \frac{2\left(e^{4x}-2+e^{-4x}-e^{4x}-2-e^{-4x}\right)}{{\left(e^{2x}-e^{-2x}\right)}^2}\\ & =& \frac{2\left(-2-2\right)}{{\left(e^{2x}-e^{-2x}\right)}^2}\\ & =& \frac{-8}{{\left(e^{2x}-e^{-2x}\right)}^2}\end{array}$

Hence, the correct option is (A).