Question

question find the derivative of $h(x)=(-x^{2}-3)^{2}sqrt{x^{2}+1}$. provide your answer below: $h(x)=square$

Explanation:

Step1: Use the quotient - rule

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

Step2: Find $u^{\prime}(x)$

First, expand $u(x)=x^{4}+6x^{2}+9$. Then $u^{\prime}(x)=4x^{3}+12x$.

Step3: Find $v^{\prime}(x)$

Using the chain - rule, if $y=(x^{2}+1)^{\frac{1}{2}}$, let $t = x^{2}+1$, then $y=t^{\frac{1}{2}}$. $\frac{dy}{dt}=\frac{1}{2}t^{-\frac{1}{2}}$ and $\frac{dt}{dx}=2x$. So $v^{\prime}(x)=\frac{1}{2}(x^{2}+1)^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{x^{2}+1}}$.

Step4: Apply the quotient - rule

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Answer:

$\frac{3x^{5}+10x^{3}+3x}{(x^{2}+1)^{\frac{3}{2}}}$