using the definition, calculate the derivative of the function. then find the values of the derivative as…

Explanation:

Step1: Recall the definition of the derivative

The derivative of a function (y = f(x)) is defined as (f^{\prime}(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}). Given (f(x)=7 - x^{2}), then (f(x + h)=7-(x + h)^{2}=7-(x^{2}+2xh+h^{2})=7 - x^{2}-2xh - h^{2}). So (\frac{f(x + h)-f(x)}{h}=\frac{(7 - x^{2}-2xh - h^{2})-(7 - x^{2})}{h}).

Step2: Simplify the expression

[ \begin{align*} \frac{(7 - x^{2}-2xh - h^{2})-(7 - x^{2})}{h}&=\frac{7 - x^{2}-2xh - h^{2}-7 + x^{2}}{h}\ &=\frac{-2xh - h^{2}}{h}\ &=\frac{h(-2x - h)}{h}\ &=-2x - h \end{align*} ]

Step3: Find the limit as (h\rightarrow0)

(f^{\prime}(x)=\lim_{h\rightarrow0}(-2x - h)). As (h\rightarrow0), we have (f^{\prime}(x)=-2x).

Step4: Calculate (f^{\prime}(-6)), (f^{\prime}(0)) and (f^{\prime}(5))

When (x=-6), (f^{\prime}(-6)=-2\times(-6) = 12). When (x = 0), (f^{\prime}(0)=-2\times0=0). When (x = 5), (f^{\prime}(5)=-2\times5=-10).

Answer:

(f^{\prime}(x)=-2x), (f^{\prime}(-6) = 12), (f^{\prime}(0)=0), (f^{\prime}(5)=-10)