if $y = xsin x$, then $\frac{dy}{dx}=$\n(a) $sin x+cos x$\n(b) $sin x + xcos x$\n(c) $sin x - xcos x$\n(d)…
if $y = xsin x$, then $\frac{dy}{dx}=$\n(a) $sin x+cos x$\n(b) $sin x + xcos x$\n(c) $sin x - xcos x$\n(d) $x(sin x+cos x)$
Answer
Explanation:
Step1: Apply product - rule
The product - rule states that if $y = u\cdot v$, then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$. Here, $u = x$ and $v=\sin x$.
Step2: Find derivatives of $u$ and $v$
The derivative of $u = x$ with respect to $x$ is $\frac{du}{dx}=1$, and the derivative of $v=\sin x$ with respect to $x$ is $\frac{dv}{dx}=\cos x$.
Step3: Substitute into product - rule
$\frac{dy}{dx}=x\cdot\cos x+\sin x\cdot1=\sin x + x\cos x$.
Answer:
B. $\sin x + x\cos x$