if (f(x)=e^{x}sin x), then (f(x)=) \na (e^{x}cos x) \nb (-e^{x}cos x) \nc (e^{x}(sin x+cos x)) \nd…
if (f(x)=e^{x}sin x), then (f(x)=) \na (e^{x}cos x) \nb (-e^{x}cos x) \nc (e^{x}(sin x+cos x)) \nd (e^{x}(sin x - cos x))
Answer
Explanation:
Step1: Recall product - rule
The product - rule states that if $y = u(x)v(x)$, then $y^\prime=u^\prime(x)v(x)+u(x)v^\prime(x)$. Here, $u(x)=e^{x}$ and $v(x)=\sin x$.
Step2: Find derivatives of $u(x)$ and $v(x)$
The derivative of $u(x) = e^{x}$ is $u^\prime(x)=e^{x}$, and the derivative of $v(x)=\sin x$ is $v^\prime(x)=\cos x$.
Step3: Apply the product - rule
$f^\prime(x)=u^\prime(x)v(x)+u(x)v^\prime(x)=e^{x}\sin x+e^{x}\cos x=e^{x}(\sin x + \cos x)$.
Answer:
C. $e^{x}(\sin x+\cos x)$