find the derivative of the following function. y = (sin x + cos x)/e^x dy/dx = □

Answer:

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

Explanation:

Step1: Apply quotient - rule

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

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

Step4: Substitute into quotient - rule

$y^\prime=\frac{(\cos x - \sin x)\mathrm{e}^{x}-(\sin x+\cos x)\mathrm{e}^{x}}{(\mathrm{e}^{x})^{2}}$.

Step5: Simplify the expression

First, factor out $\mathrm{e}^{x}$ in the numerator: $y^\prime=\frac{\mathrm{e}^{x}(\cos x - \sin x-\sin x - \cos x)}{\mathrm{e}^{2x}}$. Then, cancel out $\mathrm{e}^{x}$ from the numerator and denominator: $y^\prime=\frac{- 2\sin x}{\mathrm{e}^{x}}=\frac{-\sin x - \cos x}{\mathrm{e}^{x}}$.