Answer
$\DeclareMathOperator{\sech}{sech}$$\dfrac{d}{dx}\arctan(\tanh x)=\sech{2x}$
Work Step by Step
Firstly solve the left hand side. Thus, $\DeclareMathOperator{\sech}{sech}$$\dfrac{d}{dx}\arctan(\tanh x)=\dfrac{1}{1+\tanh^2 x}\times\sech^2 x$ As we know that $\sech^2 x=1-\tanh^2 x$
Thus, $=\dfrac{1-\tanh^2 x}{1+\tanh^2 x}$
or, $=\dfrac{\cosh^2 x-\sinh^2 x}{\cosh^2x+\sinh^2 x}$
Also, $\cosh^2 x-\sinh^2 x=1$ and $\cosh(2x)=\cosh^2 x+\sinh^2 x$
Therefore, $=\dfrac{1}{\cosh {2x}}$ $=\sech{2x}$
Hence, $\dfrac{d}{dx}\arctan(\tanh x)=\sech{2x}$