question 13 of 50 2 points use the definition of the derivative to find the derivative of each function…

Explanation:

Step1: Recall derivative definition

The definition of the derivative of a function $y = h(x)$ is $h'(x)=\lim_{h\rightarrow0}\frac{h(x + h)-h(x)}{h}$. Given $h(x)=-3x^{2}+3x + 5$, then $h(x + h)=-3(x + h)^{2}+3(x + h)+5$.

Step2: Expand $h(x + h)$

Expand $-3(x + h)^{2}+3(x + h)+5$: [ \begin{align*} -3(x + h)^{2}+3(x + h)+5&=-3(x^{2}+2xh+h^{2})+3x + 3h+5\ &=-3x^{2}-6xh-3h^{2}+3x + 3h+5 \end{align*} ]

Step3: Calculate $h(x + h)-h(x)$

[ \begin{align*} h(x + h)-h(x)&=(-3x^{2}-6xh-3h^{2}+3x + 3h+5)-(-3x^{2}+3x + 5)\ &=-3x^{2}-6xh-3h^{2}+3x + 3h+5 + 3x^{2}-3x - 5\ &=-6xh-3h^{2}+3h \end{align*} ]

Step4: Calculate $\frac{h(x + h)-h(x)}{h}$

[ \frac{h(x + h)-h(x)}{h}=\frac{-6xh-3h^{2}+3h}{h}=-6x-3h + 3 ]

Step5: Find the limit as $h\rightarrow0$

[ h'(x)=\lim_{h\rightarrow0}\frac{h(x + h)-h(x)}{h}=\lim_{h\rightarrow0}(-6x-3h + 3)=-6x+3 ]

Answer:

A. $h'(x)=-6x + 3$