if (f(x)=e^{x}sin x), then (f(x)=) \n(a) (e^{x}cos x) \n(b) (-e^{x}cos x) \n(c) (e^{x}(sin x+cos x)) \n(d)…
if (f(x)=e^{x}sin x), then (f(x)=) \n(a) (e^{x}cos x) \n(b) (-e^{x}cos x) \n(c) (e^{x}(sin x+cos x)) \n(d) (e^{x}(sin x - cos x))
Answer
Explanation:
Step1: Apply product - rule
The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Here, $u = e^{x}$ and $v=\sin x$.
Step2: Find derivatives of $u$ and $v$
The derivative of $u = e^{x}$ is $u'=e^{x}$, and the derivative of $v=\sin x$ is $v'=\cos x$.
Step3: Substitute into product - rule
$f'(x)=e^{x}\cdot\sin x+e^{x}\cdot\cos x=e^{x}(\sin x + \cos x)$
Answer:
C. $e^{x}(\sin x+\cos x)$