# How do you find the derivative of sinx/e^x?

Feb 8, 2017

$\frac{\cos x - \sin x}{e} ^ x$

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{quotient rule}}$

$\text{Given "f(x)=(g(x))/(h(x))" then}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{2}{2}} |}}} \to \left(A\right)$

$\text{here } f \left(x\right) = \frac{\sin x}{e} ^ x$

$g \left(x\right) = \sin x \Rightarrow g ' \left(x\right) = \cos x$

$\text{and } h \left(x\right) = {e}^{x} \Rightarrow h ' \left(x\right) = {e}^{x}$

Substituting these values into (A)

$f ' \left(x\right) = \frac{{e}^{x} \left(\cos x\right) - \sin x \left({e}^{x}\right)}{e} ^ \left(2 x\right)$

$\Rightarrow f ' \left(x\right) = \frac{{e}^{x} \left(\cos x - \sin x\right)}{e} ^ \left(2 x\right) = \frac{\cos x - \sin x}{e} ^ x$