Calculus: Early Transcendentals 8th Edition

LHS $=\frac{d}{dx}\arctan(\tanh x)$ $=\sech {2x}$ $=$RHS (as needed)
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)