Question

find the derivative of the following function.
y = \frac{\sin x+\cos x}{e^{x}}
\frac{dy}{dx}=\square

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = \sin x+\cos x$ and $v = e^{x}$.

Step2: Find $u^\prime$

Differentiate $u=\sin x+\cos x$ with respect to $x$. Using the sum - rule of differentiation and the derivatives of $\sin x$ and $\cos x$, we have $u^\prime=\cos x-\sin x$.

Step3: Find $v^\prime$

Differentiate $v = e^{x}$ with respect to $x$. The derivative of $e^{x}$ is $e^{x}$, so $v^\prime=e^{x}$.

Step4: Apply quotient - rule

Substitute $u$, $u^\prime$, $v$, and $v^\prime$ into the quotient - rule formula:
\[

$$\begin{align*} \frac{dy}{dx}&=\frac{(\cos x-\sin x)e^{x}-(\sin x + \cos x)e^{x}}{(e^{x})^{2}}\\ &=\frac{e^{x}(\cos x-\sin x-\sin x - \cos x)}{e^{2x}}\\ &=\frac{- 2\sin x}{e^{x}} \end{align*}$$

\]

Answer:

$\frac{-2\sin x}{e^{x}}$