Question

question
find the derivative of (h(x)=(3x^{2}+3)^{2}sqrt{x^{2}+2}).
provide your answer below:
h(x)=□

Explanation:

Step1: Use the product - rule $(uv)' = u'v+uv'$

Let $u=(3x^{2}+3)^{2}$ and $v = \sqrt{x^{2}+2}=(x^{2}+2)^{\frac{1}{2}}$.

Step2: Find $u'$ using the chain - rule

If $y = u^{2}$ with $u = 3x^{2}+3$, then $\frac{dy}{du}=2u$ and $\frac{du}{dx}=6x$. So $u'=\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=2(3x^{2}+3)\cdot6x = 12x(3x^{2}+3)$.

Step3: Find $v'$ using the chain - rule

If $y = v^{\frac{1}{2}}$ with $v=x^{2}+2$, then $\frac{dy}{dv}=\frac{1}{2}v^{-\frac{1}{2}}$ and $\frac{dv}{dx}=2x$. So $v'=\frac{dy}{dx}=\frac{dy}{dv}\cdot\frac{dv}{dx}=\frac{1}{2}(x^{2}+2)^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{x^{2}+2}}$.

Step4: Apply the product - rule

$h'(x)=u'v + uv'=12x(3x^{2}+3)\sqrt{x^{2}+2}+(3x^{2}+3)^{2}\frac{x}{\sqrt{x^{2}+2}}$.
Simplify the expression:
\[

$$\begin{align*} h'(x)&=\frac{12x(3x^{2}+3)(x^{2}+2)+(3x^{2}+3)^{2}x}{\sqrt{x^{2}+2}}\\ &=\frac{x(3x^{2}+3)[12(x^{2}+2)+(3x^{2}+3)]}{\sqrt{x^{2}+2}}\\ &=\frac{x(3x^{2}+3)(12x^{2}+24 + 3x^{2}+3)}{\sqrt{x^{2}+2}}\\ &=\frac{x(3x^{2}+3)(15x^{2}+27)}{\sqrt{x^{2}+2}}\\ &=\frac{3x(3x^{2}+3)(5x^{2}+9)}{\sqrt{x^{2}+2}} \end{align*}$$

\]

Answer:

$\frac{3x(3x^{2}+3)(5x^{2}+9)}{\sqrt{x^{2}+2}}$