Question

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

Explanation:

Step1: Use the product - rule

The product - rule states that if $h(x)=u(x)v(x)$, then $h^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$. Let $u(x)=(2 - 3x^{2})^{3}$ and $v(x)=\sqrt{2x^{2}+2}=(2x^{2}+2)^{\frac{1}{2}}$.

Step2: Find $u^{\prime}(x)$ using the chain - rule

The chain - rule states that if $y = f(g(x))$, then $y^{\prime}=f^{\prime}(g(x))g^{\prime}(x)$. For $u(x)=(2 - 3x^{2})^{3}$, let $t = 2-3x^{2}$, so $u(t)=t^{3}$. Then $u^{\prime}(t) = 3t^{2}$ and $t^{\prime}=-6x$. So $u^{\prime}(x)=3(2 - 3x^{2})^{2}\times(-6x)=-18x(2 - 3x^{2})^{2}$.

Step3: Find $v^{\prime}(x)$ using the chain - rule

For $v(x)=(2x^{2}+2)^{\frac{1}{2}}$, let $s = 2x^{2}+2$, so $v(s)=s^{\frac{1}{2}}$. Then $v^{\prime}(s)=\frac{1}{2}s^{-\frac{1}{2}}$ and $s^{\prime}=4x$. So $v^{\prime}(x)=\frac{1}{2}(2x^{2}+2)^{-\frac{1}{2}}\times4x=\frac{2x}{\sqrt{2x^{2}+2}}$.

Step4: Apply the product - rule

$h^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$
$h^{\prime}(x)=-18x(2 - 3x^{2})^{2}\sqrt{2x^{2}+2}+(2 - 3x^{2})^{3}\frac{2x}{\sqrt{2x^{2}+2}}$
$h^{\prime}(x)=\frac{-18x(2 - 3x^{2})^{2}(2x^{2}+2)+2x(2 - 3x^{2})^{3}}{\sqrt{2x^{2}+2}}$
$h^{\prime}(x)=\frac{2x(2 - 3x^{2})^{2}[-9(2x^{2}+2)+(2 - 3x^{2})]}{\sqrt{2x^{2}+2}}$
$h^{\prime}(x)=\frac{2x(2 - 3x^{2})^{2}(-18x^{2}-18 + 2-3x^{2})}{\sqrt{2x^{2}+2}}$
$h^{\prime}(x)=\frac{2x(2 - 3x^{2})^{2}(-21x^{2}-16)}{\sqrt{2x^{2}+2}}$

Answer:

$\frac{2x(2 - 3x^{2})^{2}(-21x^{2}-16)}{\sqrt{2x^{2}+2}}$