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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 $\tanh x=\frac{\sinh x}{\cosh x}$.

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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Still her you to the two x plus two e to the x into the negative x plus any to the negative to X No numerator is gonna be the same thing. Soap. Its attraction so X minus y squared equals X squared minus two. Ex wife close wars earlier So it's essentially the same thing. But you have reminders to ex mines that positive Well, in a perfect stone here, so four plus e to the two x my last two e to the X seats of the negative x plus e to the negative to X. So now, as we said before, next to the A terms accidentally equals X to the A plus B So I'm gonna use it to cancel out this heart right here. This each of the works and it's it's a negative X that will become one one because one minus one is zero and each of the zero power Anything to zero powers. One Mr. Everything that happened down here to this part right here. Don't counsel out because it's one and I will continue to simple for expression. 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Mississippi State University

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