Question

question
find the derivative of h(x)=(3x^2 + 1)^2√(x^3+1).
provide your answer below:
h(x)=□

Explanation:

Step1: Use product - rule

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

Step2: Find \(u^\prime\)

Use the chain - rule. If \(y = f(g(x))\), then \(y^\prime=f^\prime(g(x))\cdot g^\prime(x)\). For \(u=(3x^{2}+1)^{2}\), let \(t = 3x^{2}+1\), so \(u = t^{2}\). Then \(u^\prime=\frac{du}{dt}\cdot\frac{dt}{dx}\). \(\frac{du}{dt}=2t\) and \(\frac{dt}{dx}=6x\). Substituting \(t = 3x^{2}+1\) back, we get \(u^\prime=2(3x^{2}+1)\cdot6x = 12x(3x^{2}+1)\).

Step3: Find \(v^\prime\)

For \(v=(x^{2}+1)^{\frac{1}{2}}\), using the chain - rule. Let \(s=x^{2}+1\), so \(v = s^{\frac{1}{2}}\). Then \(v^\prime=\frac{dv}{ds}\cdot\frac{ds}{dx}\). \(\frac{dv}{ds}=\frac{1}{2}s^{-\frac{1}{2}}\) and \(\frac{ds}{dx}=2x\). Substituting \(s = x^{2}+1\) back, we get \(v^\prime=\frac{1}{2}(x^{2}+1)^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{x^{2}+1}}\).

Step4: Apply the product - rule

\(h^\prime(x)=u^\prime v+uv^\prime\).
\[

$$\begin{align*} h^\prime(x)&=12x(3x^{2}+1)\sqrt{x^{2}+1}+(3x^{2}+1)^{2}\cdot\frac{x}{\sqrt{x^{2}+1}}\\ &=\frac{12x(3x^{2}+1)(x^{2}+1)+(3x^{2}+1)^{2}x}{\sqrt{x^{2}+1}}\\ &=\frac{x(3x^{2}+1)[12(x^{2}+1)+(3x^{2}+1)]}{\sqrt{x^{2}+1}}\\ &=\frac{x(3x^{2}+1)(12x^{2}+12 + 3x^{2}+1)}{\sqrt{x^{2}+1}}\\ &=\frac{x(3x^{2}+1)(15x^{2}+13)}{\sqrt{x^{2}+1}} \end{align*}$$

\]

Answer:

\(\frac{x(3x^{2}+1)(15x^{2}+13)}{\sqrt{x^{2}+1}}\)