find the derivative of the function.\nh(t)=(t + 5)^{2/3}(2t^{2}-1)^{3}\nh(t)=□

Explanation:

Step1: Apply product - rule

The product - rule states that if $h(t)=u(t)v(t)$, then $h^{\prime}(t)=u^{\prime}(t)v(t)+u(t)v^{\prime}(t)$. Let $u(t)=(t + 5)^{\frac{2}{3}}$ and $v(t)=(2t^{2}-1)^{3}$.

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

The chain - rule: if $y = f(g(t))$, then $y^{\prime}=f^{\prime}(g(t))g^{\prime}(t)$. For $u(t)=(t + 5)^{\frac{2}{3}}$, let $g(t)=t + 5$ and $f(x)=x^{\frac{2}{3}}$. Then $u^{\prime}(t)=\frac{2}{3}(t + 5)^{-\frac{1}{3}}\times1=\frac{2}{3(t + 5)^{\frac{1}{3}}}$.

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

For $v(t)=(2t^{2}-1)^{3}$, let $g(t)=2t^{2}-1$ and $f(x)=x^{3}$. Then $v^{\prime}(t)=3(2t^{2}-1)^{2}\times4t = 12t(2t^{2}-1)^{2}$.

Step4: Calculate $h^{\prime}(t)$

$h^{\prime}(t)=u^{\prime}(t)v(t)+u(t)v^{\prime}(t)=\frac{2}{3(t + 5)^{\frac{1}{3}}}(2t^{2}-1)^{3}+(t + 5)^{\frac{2}{3}}\times12t(2t^{2}-1)^{2}$. Factor out $(t + 5)^{-\frac{1}{3}}(2t^{2}-1)^{2}$: [ \begin{align*} h^{\prime}(t)&=(t + 5)^{-\frac{1}{3}}(2t^{2}-1)^{2}\left[\frac{2(2t^{2}-1)}{3}+12t(t + 5)\right]\ &=\frac{(2t^{2}-1)^{2}}{(t + 5)^{\frac{1}{3}}}\left(\frac{4t^{2}-2}{3}+12t^{2}+60t\right)\ &=\frac{(2t^{2}-1)^{2}}{(t + 5)^{\frac{1}{3}}}\left(\frac{4t^{2}-2 + 36t^{2}+180t}{3}\right)\ &=\frac{(2t^{2}-1)^{2}(40t^{2}+180t - 2)}{3(t + 5)^{\frac{1}{3}}} \end{align*} ]

Answer:

$\frac{(2t^{2}-1)^{2}(40t^{2}+180t - 2)}{3(t + 5)^{\frac{1}{3}}}$