Question

question find the derivative of (h(x)=(3 - 2x^{2})^{3}sqrt{2x^{2}+1}). 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 \(u(x)=(3 - 2x^{2})^{3}\) and \(v(x)=\sqrt{2x^{2}+1}=(2x^{2}+1)^{\frac{1}{2}}\).

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 \(u(x)=(3 - 2x^{2})^{3}\), let \(t = 3-2x^{2}\), so \(u(x)=t^{3}\). Then \(\frac{du}{dt}=3t^{2}=3(3 - 2x^{2})^{2}\) and \(\frac{dt}{dx}=-4x\). So \(u^{\prime}(x)=\frac{du}{dt}\cdot\frac{dt}{dx}=3(3 - 2x^{2})^{2}\cdot(-4x)=-12x(3 - 2x^{2})^{2}\).

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

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

Step4: Apply the product - rule

\[

$$\begin{align*} h^{\prime}(x)&=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)\\ &=-12x(3 - 2x^{2})^{2}\sqrt{2x^{2}+1}+(3 - 2x^{2})^{3}\cdot\frac{2x}{\sqrt{2x^{2}+1}}\\ &=\frac{-12x(3 - 2x^{2})^{2}(2x^{2}+1)+2x(3 - 2x^{2})^{3}}{\sqrt{2x^{2}+1}}\\ &=\frac{2x(3 - 2x^{2})^{2}[-6(2x^{2}+1)+(3 - 2x^{2})]}{\sqrt{2x^{2}+1}}\\ &=\frac{2x(3 - 2x^{2})^{2}(-12x^{2}-6 + 3-2x^{2})}{\sqrt{2x^{2}+1}}\\ &=\frac{2x(3 - 2x^{2})^{2}(-14x^{2}-3)}{\sqrt{2x^{2}+1}}\\ &=-\frac{2x(3 - 2x^{2})^{2}(14x^{2}+3)}{\sqrt{2x^{2}+1}} \end{align*}$$

\]

Answer:

\(-\frac{2x(3 - 2x^{2})^{2}(14x^{2}+3)}{\sqrt{2x^{2}+1}}\)