Question

question
find $f(x)$ where $f(x)=-\frac{1}{2}x^{3}cos(x)$.
provide your answer below:
$f(x)=square$

Explanation:

Step1: Recall product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Here, $u =-\frac{1}{2}x^{3}$ and $v=\cos(x)$.

Step2: Differentiate $u$

Differentiate $u =-\frac{1}{2}x^{3}$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $u'=-\frac{1}{2}\times3x^{2}=-\frac{3}{2}x^{2}$.

Step3: Differentiate $v$

Differentiate $v=\cos(x)$ with respect to $x$. We know that $\frac{d}{dx}(\cos(x))=-\sin(x)$.

Step4: Apply product - rule

$f'(x)=u'v + uv'$. Substitute $u =-\frac{1}{2}x^{3}$, $u'=-\frac{3}{2}x^{2}$, $v=\cos(x)$ and $v'=-\sin(x)$ into the formula.
$f'(x)=-\frac{3}{2}x^{2}\cos(x)-\frac{1}{2}x^{3}(-\sin(x))$.
Simplify the expression: $f'(x)=-\frac{3}{2}x^{2}\cos(x)+\frac{1}{2}x^{3}\sin(x)$.

Answer:

$-\frac{3}{2}x^{2}\cos(x)+\frac{1}{2}x^{3}\sin(x)$