Question

2 mark for review if $f(x)=\frac{sin x}{e^{x}}$, then $f(x)=$ a $\frac{-cos x - sin x}{e^{x}}$ b $\frac{cos x-sin x}{e^{x}}$ c $\frac{sin x - cos x}{e^{x}}$ d $\frac{cos x+sin x}{e^{x}}$

Explanation:

Step1: Recall quotient - rule

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

Step2: Apply quotient - rule

$f'(x)=\frac{(\sin x)'\cdot e^{x}-\sin x\cdot(e^{x})'}{(e^{x})^{2}}$. Substituting $(\sin x)'=\cos x$ and $(e^{x})' = e^{x}$ into the formula, we get $f'(x)=\frac{\cos x\cdot e^{x}-\sin x\cdot e^{x}}{(e^{x})^{2}}$.

Step3: Simplify the expression

Factor out $e^{x}$ from the numerator: $f'(x)=\frac{e^{x}(\cos x-\sin x)}{e^{2x}}$. Then cancel out one factor of $e^{x}$ in the numerator and denominator, resulting in $f'(x)=\frac{\cos x - \sin x}{e^{x}}$.

Answer:

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