Question

question. find the derivative of (h(x)=(2 - 3x^{2})^{2}sqrt{2x^{2}+2}). (h(x)=4 - 12x^{2}+9x^{4}) sorry, thats incorrect. try again?

Explanation:

Step1: Rewrite the function

First, rewrite $h(x)=(2 - 3x^{2})^{2}\sqrt{2x^{2}+2}$. Let $u=(2 - 3x^{2})^{2}$ and $v = \sqrt{2x^{2}+2}=(2x^{2}+2)^{\frac{1}{2}}$. Then $h(x)=u\cdot v$.

Step2: Find the derivative of $u$ using the chain - rule

If $y = u^{2}$ where $u = 2-3x^{2}$, then $\frac{dy}{du}=2u$ and $\frac{du}{dx}=-6x$. So, $\frac{du}{dx}=2(2 - 3x^{2})\cdot(-6x)=-12x(2 - 3x^{2})=-24x + 36x^{3}$.

Step3: Find the derivative of $v$ using the chain - rule

If $y = v^{\frac{1}{2}}$ where $v = 2x^{2}+2$, then $\frac{dy}{dv}=\frac{1}{2}v^{-\frac{1}{2}}$ and $\frac{dv}{dx}=4x$. So, $\frac{dv}{dx}=\frac{1}{2}(2x^{2}+2)^{-\frac{1}{2}}\cdot4x=\frac{2x}{\sqrt{2x^{2}+2}}$.

Step4: Use the product - rule

The product - rule states that $(uv)^\prime=u^\prime v+uv^\prime$.
\[

$$\begin{align*} h^\prime(x)&=(-24x + 36x^{3})\sqrt{2x^{2}+2}+(2 - 3x^{2})^{2}\cdot\frac{2x}{\sqrt{2x^{2}+2}}\\ &=\frac{(-24x + 36x^{3})(2x^{2}+2)+2x(2 - 3x^{2})^{2}}{\sqrt{2x^{2}+2}}\\ &=\frac{-48x^{3}-48x + 72x^{5}+72x^{3}+2x(4 - 12x^{2}+9x^{4})}{\sqrt{2x^{2}+2}}\\ &=\frac{-48x^{3}-48x + 72x^{5}+72x^{3}+8x-24x^{3}+18x^{5}}{\sqrt{2x^{2}+2}}\\ &=\frac{(72x^{5}+18x^{5})+(-48x^{3}+72x^{3}-24x^{3})+(-48x + 8x)}{\sqrt{2x^{2}+2}}\\ &=\frac{90x^{5}-40x}{\sqrt{2x^{2}+2}} \end{align*}$$

\]

Answer:

$\frac{90x^{5}-40x}{\sqrt{2x^{2}+2}}$