Question

given the function $f(x)=(x^{3}-3x + 5)cos x$, find $f(x)$ in any form.

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u=x^{3}-3x + 5$ and $v = \cos x$.

Step2: Differentiate $u$

Differentiate $u=x^{3}-3x + 5$ with respect to $x$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we get $u'=\frac{d}{dx}(x^{3}-3x + 5)=3x^{2}-3$.

Step3: Differentiate $v$

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

Step4: Substitute $u$, $u'$, $v$, and $v'$ into product - rule

$y'=u'v+uv'=(3x^{2}-3)\cos x+(x^{3}-3x + 5)(-\sin x)$.
Simplify to get $y'=(3x^{2}-3)\cos x-(x^{3}-3x + 5)\sin x$.

Answer:

$(3x^{2}-3)\cos x-(x^{3}-3x + 5)\sin x$