Answer
a) The derivative of $f\left( x \right)={{x}^{2}}-8$ at x is $f'\left( x \right)=2x$.
b) The slope of the tangent line to the graph of $f\left( x \right)={{x}^{2}}-8$ at $x=-1$ is $f'\left( -1 \right)=-2$ and at $x=3$ is $f'\left( 3 \right)=6$.
Work Step by Step
(a)
Consider the function, $f\left( x \right)={{x}^{2}}-8$
Now, compute the derivate of $f\left( x \right)={{x}^{2}}-8$ using the formula $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$
To compute $f\left( x+h \right)$, substitute $x=x+h$ in the function $f\left( x \right)={{x}^{2}}-8$.
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left[ {{\left( x+h \right)}^{2}}-8 \right]-\left( {{x}^{2}}-8 \right)}{h}$
Now, simplify ${{\left( x+h \right)}^{2}}$ by using the property ${{\left( A+B \right)}^{2}}={{A}^{2}}+2AB+{{B}^{2}}$.
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}+2xh+{{h}^{2}}-8-{{x}^{2}}+8}{h}$
Combine the like terms in the numerator; this gives,
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{2xh+{{h}^{2}}}{h}$
Divide numerator and denominator by h.
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( 2x+h \right)$
Apply the limits,
$\begin{align}
& f'\left( x \right)=2x+0 \\
& =2x
\end{align}$
Thus, the derivative of $f\left( x \right)={{x}^{2}}-8$ at x is $f'\left( x \right)=2x$.
(b)
Consider the function, $f\left( x \right)={{x}^{2}}-8$
From part (a), the derivative of $f\left( x \right)={{x}^{2}}-8$ at x is $f'\left( x \right)=2x$.
To compute the slope of the tangent line at $x=-1$, substitute $x=-1$ in $f'\left( x \right)$.
$\begin{align}
& f'\left( -1 \right)=2\left( -1 \right) \\
& =-2
\end{align}$
To compute the slope of the tangent line at $x=3$, substitute $x=3$ in $f'\left( x \right)$.
$\begin{align}
& f'\left( 3 \right)=2\left( 3 \right) \\
& =6
\end{align}$
