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Problem 17 Easy Difficulty

The hyperbolic functions are defined as $\sinh x=\frac{1}{2}\left(e^{x}-e^{-x}\right)$ and $\cosh x=\frac{1}{2}\left(e^{x}+e^{-x}\right)$.
a. Prove $\frac{d(\sinh x)}{d x}=\cosh x$.
b. Prove $\frac{d(\cosh x)}{d x}=\sinh x$.
c. Prove $\frac{d(\tanh x)}{d x}=\frac{1}{(\cosh x)^{2}}$ if $\tanh x=\frac{\sinh x}{\cosh x}$.

Answer

a. $\frac{d}{d x}(\sinh x)=\frac{d}{d x}\left[\frac{1}{2}\left(e^{x}-e^{-x}\right)\right]$
$$
\begin{array}{l}
=\frac{1}{2}\left(e^{x}+e^{-x}\right) \\
=\cosh x
\end{array}
$$
b. $\frac{d}{d x}(\cosh x)=\frac{1}{2}\left(e^{t}-e^{-t}\right)$
$=\sinh x$
c. Since tanh $x=\frac{\sinh x}{\cosh x^{\prime}}$
$\begin{aligned} \frac{d}{d x}(\tanh x)=\frac{\left(\frac{d}{d x} \sinh x\right)(\cosh x)}{(\cosh x)^{2}} \\-\frac{(\sinh x)\left(\frac{d}{d x} \cosh x\right)}{(\cosh x)^{2}} \\=& \frac{\frac{1}{2}\left(e^{x}+e^{-x}\right)\left(\frac{1}{2}\right)\left(e^{x}+e^{-x}\right)}{(\cosh x)^{2}} \\=& \frac{\frac{1}{4}(4)}{(\cosh x)^{2}} \\=& \frac{1}{(\cosh x)^{2}} \end{aligned}$

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Video Transcript

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