Answer
LHS
$=\frac{d}{dx}\arctan(\tanh x)$
$=\sech {2x}$
$=$RHS (as needed)
Work Step by Step
LHS
$=\frac{d}{dx}\arctan(\tanh x)$
Using the chain rule:
LHS
$=\frac{d\arctan(\tanh x)}{d\tanh x}
\times\frac{d\tanh x}{dx}$
$\DeclareMathOperator{\sech}{sech}$
$=\frac{1}{1+\tanh^2 x}\times\sech^2 x$
Recall that:
$$\sech^2 x=1-\tanh^2 x$$
LHS
$=\frac{1-\tanh^2 x}{1+\tanh^2 x}$
Multiply the numerator and denominator by $\cosh^2x$:
LHS
$=\frac{\cosh^2 x-\sinh^2 x}{\cosh^2x+\sinh^2 x}$
Recall that:
1) $$\cosh^2 x-\sinh^2 x=1$$
2) $$\cosh(2x)=\cosh^2 x+\sinh^2 x$$ (derived from the compound formula for $\cosh$)
Thus,
LHS
$=\frac{1}{\cosh {2x}}$
$=\sech{2x}$
$=$RHS (as needed)
