Answer
See the proof below.
Work Step by Step
Using the definition of $\tanh x$ and the quotient rule, we have
\begin{align*} \frac{d}{d x} \tanh x &=\frac{d}{d x}\left(\frac{\sinh x}{\cosh x}\right) \\ &=\frac{\cosh x \cdot \sinh x^{\prime}-\sinh x \cdot \cosh ^{\prime} x}{(\cosh x)^{2}} \\ &=\frac{\cosh x \cdot \cosh x-\sinh x \cdot \sinh x}{\cosh ^{2} x} \\
&=\frac{\cosh ^{2} x-\sinh ^{2} x}{\cosh ^{2} x} \\
&=1-\tanh ^{2} x \\
&=\operatorname{sech}^{2} x \end{align*}
where we used the fact that $\left(\tanh ^{2} x+\operatorname{sech}^{2} x=1\right) .$
