Answer
LHS
$=\frac{d}{dx}_4\sqrt{\frac{1+\tanh x}{1-\tanh x}}$
$=\frac{1}{2}{e^{\frac{x}{2}}}$
=RHS (as needed)
Work Step by Step
LHS
$=\frac{d}{dx}_4\sqrt{\frac{1+\tanh x}{1-\tanh x}}$
$=\frac{d}{dx}(\frac{1+\tanh x}{1-\tanh x})^{\frac{1}{4}}$
Using the chain rule:
LHS
$=\frac{d(\frac{1+\tanh x}{1-\tanh x})^{\frac{1}{4}}}{d\frac{1+\tanh x}{1-\tanh x}}
\times\frac{d\frac{1+\tanh x}{1-\tanh x}}{dx}$
$=\frac{1}{4(\frac{1+\tanh x}{1-\tanh x})^{\frac{3}{4}}}
\times\frac{d\frac{1+\tanh x}{1-\tanh x}}{dx}$
$\DeclareMathOperator{\sech}{sech}$
Using the quotient rule:
LHS
$=\frac{1}{4(\frac{1+\tanh x}{1-\tanh x})^{\frac{3}{4}}}
\times\frac{\sech^2x(1-\tanh x)-(-\sech^2x)(1+\tanh x)}{(1-\tanh x)^2}$
$=\frac{1}{4(\frac{1+\tanh x}{1-\tanh x})^{\frac{3}{4}}}
\times\frac{\sech^2x-\sech^2x\tanh x+\sech^2x+\sech^2x\tanh x}{(1-\tanh x)^2}$
$=\frac{1}{4(\frac{1+\tanh x}{1-\tanh x})^{\frac{3}{4}}}
\times\frac{2\sech^2x}{(1-\tanh x)^2}$
Recall that:
$$\sech^2 x=1-\tanh^2 x$$
Thus,
LHS
$=\frac{1}{2(\frac{1+\tanh x}{1-\tanh x})^{\frac{3}{4}}}
\times\frac{(1-\tanh x)(1+\tanh x)}{(1-\tanh x)^2}$
$=\frac{1}{2}(\frac{1-\tanh x}{1+\tanh x})^{\frac{3}{4}}
\times\frac{1+\tanh x}{1-\tanh x}$
$=\frac{1}{2}(\frac{1-\tanh x}{1+\tanh x})^{\frac{3}{4}}
\times\frac{1+\tanh x}{1-\tanh x}$
$=\frac{1}{2}(\frac{1+\tanh x}{1-\tanh x})^{\frac{1}{4}}$
Recall that:
$$\tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$
Hence,
LHS
$=\frac{1}{2}(\frac{1+\frac{e^x-e^{-x}}{e^x+e^{-x}}}{1-\frac{e^x-e^{-x}}{e^x+e^{-x}}})^{\frac{1}{4}}$
$=\frac{1}{2}(\frac{e^x+e^{-x}+e^x-e^{-x}}{e^x+e^{-x}-e^x+e^{-x}})^{\frac{1}{4}}$
$=\frac{1}{2}(\frac{2e^x}{2e^{-x}})^{\frac{1}{4}}$
$=\frac{1}{2}(e^{2x})^{\frac{1}{4}}$
$=\frac{1}{2}{e^{\frac{x}{2}}}$
=RHS (as needed)
