Differentiation of Hyperbolic Functions

Example 1:

$$ y=\sinh x,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

\begin{align*} y&=\sinh x \\ \frac{dy}{dx}&=\frac{d}{dx}\sinh x \\ &=\frac{d}{dx} \left(\frac{e^x-e^{-x}}{2}\right) \\ &=\frac{1}{2}\frac{d}{dx} (e^x-e^{-x}) \\ &=\frac{1}{2}\left(\frac{d}{dx} e^x-\frac{d}{dx} e^{-x}\right) \\ &=\frac{1}{2}(e^x-(-e^{-x})) \\ &=\frac{e^x+e^{-x}}{2} \\ &=\cosh x \\ \end{align*}

Example 2:

$$ y=\cosh x,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Proof:

\begin{align*} y&=\cosh x \\ \frac{dy}{dx}&=\frac{d}{dx}\cosh x \\ &=\frac{d}{dx} \left(\frac{e^x+e^{-x}}{2}\right) \\ &=\frac{1}{2}\frac{d}{dx} (e^x+e^{-x}) \\ &=\frac{1}{2}\left(\frac{d}{dx} e^x+\frac{d}{dx} e^{-x}\right) \\ &=\frac{1}{2}(e^x+(-e^{-x})) \\ &=\frac{e^x-e^{-x}}{2} \\ &=\sinh x \\ \end{align*}

Example 1:

$$ y=\sinh 5x,\;\;\; \textup{Find} \;\;\frac{dy}{dx}$$

Solution:

$$ y=\sinh 5x$$ $$ \frac{dy}{dx}=\frac{d}{dx}\sinh 5x$$ $$ \textup{Let}\; u= 5x$$ $$ \frac{dy}{dx}=\frac{d}{dx} \sinh u$$ $$ \frac{dy}{dx}=\frac{d}{du} \sinh u \frac{du}{dx}$$ $$ \because\frac{d}{dx}\sinh x=\cosh x$$ $$ \therefore\frac{dy}{dx}=\cosh u \frac{d}{dx}4x$$ $$ \frac{dy}{dx}=\cosh 5x (5\frac{d}{dx}x)$$ $$ \because\frac{d}{dx} x=1 $$ $$ \therefore \frac{dy}{dx}=\cosh 5x (5(1))$$ $$ \frac{dy}{dx}=\cosh 5x (5)$$ $$ \frac{dy}{dx}=5\cosh 5x $$