Question

question 3
find the derivative of g(x)=(x^3 + 3x + 3)^3.
○ g(x)=3(x^3 + 3x + 3)^2
○ g(x)=27(x^2 + 1)^3
○ g(x)=9x^8 + 81x^2
○ g(x)=9(x^3 + 3x + 3)^2(x^2 + 1)

Explanation:

Step1: Apply chain - rule

Let $u = x^{3}+3x + 3$, then $g(x)=u^{3}$. The chain - rule states that $\frac{dg}{dx}=\frac{dg}{du}\cdot\frac{du}{dx}$. First, find $\frac{dg}{du}$.
$\frac{dg}{du}=\frac{d}{du}(u^{3}) = 3u^{2}$

Step2: Find $\frac{du}{dx}$

$\frac{du}{dx}=\frac{d}{dx}(x^{3}+3x + 3)=3x^{2}+3 = 3(x^{2}+1)$

Step3: Calculate $\frac{dg}{dx}$

By the chain - rule $\frac{dg}{dx}=\frac{dg}{du}\cdot\frac{du}{dx}$. Substitute $u = x^{3}+3x + 3$, $\frac{dg}{du}=3u^{2}$ and $\frac{du}{dx}=3(x^{2}+1)$ into the formula.
$\frac{dg}{dx}=3(x^{3}+3x + 3)^{2}\cdot3(x^{2}+1)=9(x^{3}+3x + 3)^{2}(x^{2}+1)$

Answer:

$g^{\prime}(x)=9(x^{3}+3x + 3)^{2}(x^{2}+1)$