# What is the derivative of sinh(x)?

Dec 12, 2014

$\frac{d \left(\sinh \left(x\right)\right)}{\mathrm{dx}} = \cosh \left(x\right)$

Proof: It is helpful to note that $\sinh \left(x\right) : = \frac{{e}^{x} - {e}^{-} x}{2}$ and $\cosh \left(x\right) : = \frac{{e}^{x} + {e}^{-} x}{2}$. We can differentiate from here using either the quotient rule or the sum rule. I'll use the sum rule first:
$\sinh \left(x\right) = \frac{{e}^{x} - {e}^{-} x}{2}$
$= \frac{{e}^{x}}{2} - \frac{{e}^{-} x}{2}$
$\implies \frac{d \left(\sinh \left(x\right)\right)}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\frac{{e}^{x}}{2} - \frac{{e}^{-} x}{2}\right)$
$= \frac{d}{\mathrm{dx}} \left({e}^{x} / 2\right) - \frac{d}{\mathrm{dx}} \left({e}^{-} \frac{x}{2}\right)$ by the sum rule
$= {e}^{x} / 2 - \left(- {e}^{x} / 2\right)$ by basic differentiation of exponential functions
$= \frac{{e}^{x} + {e}^{-} x}{2} = \cosh \left(x\right) .$

The quotient rule is just as easy:
Let $u = {e}^{x} - {e}^{-} x$ and $v = 2$
Hence $\frac{\mathrm{du}}{\mathrm{dx}} = {e}^{x} + {e}^{-} x$ (by the sum rule and basic differentiation of exponential functions)
and $\frac{\mathrm{dv}}{\mathrm{dx}} = 0$
Recalling that $\frac{d \left(\frac{u}{v}\right)}{\mathrm{dx}} = \frac{v \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) - u \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right)}{v} ^ 2$
so $\frac{d \left(\sinh \left(x\right)\right)}{\mathrm{dx}} = \frac{2 \cdot \left({e}^{x} + {e}^{-} x\right) - 0}{2} ^ 2$
$= \frac{2 \cdot \left({e}^{x} + {e}^{-} x\right)}{4}$
$= \frac{{e}^{x} + {e}^{-} x}{2} = \cosh \left(x\right) .$