Question

consider the following. f(x) = x sin(x) find the first derivative of the function. f(x) = cos(x) × find the second derivative of the function. f(x) = -sin(x) ×

Explanation:

Step1: Apply product - rule for first derivative

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. For $f(x)=x\sin(x)$, let $u = x$ and $v=\sin(x)$. Then $u' = 1$ and $v'=\cos(x)$. So $f'(x)=1\cdot\sin(x)+x\cdot\cos(x)=\sin(x)+x\cos(x)$.

Step2: Apply product - rule for second derivative

We have $f'(x)=\sin(x)+x\cos(x)$. The derivative of $\sin(x)$ is $\cos(x)$. For the second term $x\cos(x)$, using the product - rule with $u = x$ and $v=\cos(x)$ (where $u' = 1$ and $v'=-\sin(x)$), the derivative of $x\cos(x)$ is $1\cdot\cos(x)+x\cdot(-\sin(x))=\cos(x)-x\sin(x)$. Then $f''(x)=\cos(x)+\cos(x)-x\sin(x)=2\cos(x)-x\sin(x)$.

Answer:

$f'(x)=\sin(x)+x\cos(x)$
$f''(x)=2\cos(x)-x\sin(x)$