question 4 of 8 step 1 of 1 find the derivative for the given function. write your answer using positive and…

Explanation:

Step1: Apply product - rule

The product - rule states that if $h(x)=u(x)v(x)$, then $h^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$. Here, $u(x)=-8x$ and $v(x)=(2x^{3}+9)^{3}$. First, find $u^{\prime}(x)$ and $v^{\prime}(x)$. $u^{\prime}(x)=-8$

Step2: Apply chain - rule to find $v^{\prime}(x)$

Let $t = 2x^{3}+9$, so $v(x)=t^{3}$. By the chain - rule $\frac{dv}{dx}=\frac{dv}{dt}\cdot\frac{dt}{dx}$. $\frac{dv}{dt}=3t^{2}=3(2x^{3}+9)^{2}$ and $\frac{dt}{dx}=6x^{2}$. Then $v^{\prime}(x)=3(2x^{3}+9)^{2}\cdot6x^{2}=18x^{2}(2x^{3}+9)^{2}$

Step3: Calculate $h^{\prime}(x)$

Using the product - rule $h^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$. $h^{\prime}(x)=-8(2x^{3}+9)^{3}+(-8x)\cdot18x^{2}(2x^{3}+9)^{2}$ $=-8(2x^{3}+9)^{3}-144x^{3}(2x^{3}+9)^{2}$

Answer:

$-8(2x^{3}+9)^{3}-144x^{3}(2x^{3}+9)^{2}$