Answer
$\dfrac{d}{dx}\sqrt[4]{\dfrac{1+\tanh x}{1-\tanh x}}=\dfrac{1}{2}{e^{\frac{x}{2}}}$
Work Step by Step
As we we know that $\tanh x=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$
Consider left hand side: $\dfrac{d}{dx}\sqrt[4]{\dfrac{1+\tanh x}{1-\tanh x}}=\dfrac{1}{2}(\dfrac{1+\dfrac{e^x-e^{-x}}{e^x+e^{-x}}}{1-\dfrac{e^x-e^{-x}}{e^x+e^{-x}}})^{\dfrac{1}{4}}$
or, $=\dfrac{1}{2}(\dfrac{e^x+e^{-x}+e^x-e^{-x}}{e^x+e^{-x}-e^x+e^{-x}})^{\frac{1}{4}}$
or, $=\dfrac{1}{2}(\dfrac{2e^x}{2e^{-x}})^{\frac{1}{4}}$
or, $=\dfrac{1}{2}(e^{2x})^{\dfrac{1}{4}}$
Hence, $\dfrac{d}{dx}\sqrt[4]{\dfrac{1+\tanh x}{1-\tanh x}}=\dfrac{1}{2}{e^{\frac{x}{2}}}$
