Question

let (f) be a differentiable function such that (f(2)=2) and (f^{prime}(2)=5). if (g(x)=x^{3}f(x)), what is the value of (g^{prime}(2))?
a 17
b 24
c 60
d 64

Explanation:

Step1: Apply product - rule

The product - rule states that if $g(x)=u(x)v(x)$, then $g^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$. Here, $u(x)=x^{3}$ and $v(x)=f(x)$. So, $g^{\prime}(x)=(x^{3})^{\prime}f(x)+x^{3}f^{\prime}(x)$.
Since $(x^{3})^{\prime}=3x^{2}$, we have $g^{\prime}(x)=3x^{2}f(x)+x^{3}f^{\prime}(x)$.

Step2: Substitute $x = 2$

Substitute $x = 2$ into $g^{\prime}(x)$.
We know that $f(2)=2$ and $f^{\prime}(2)=5$.
$g^{\prime}(2)=3\times2^{2}\times f(2)+2^{3}\times f^{\prime}(2)$.
$g^{\prime}(2)=3\times4\times2 + 8\times5$.
$g^{\prime}(2)=24 + 40$.
$g^{\prime}(2)=64$.

Answer:

D. 64