Answer
$ a)$ $General formula for Slope of tangent line =$$m_{tan}=2x_{0}$
$b)$ $Slope of tangent line =$$m_{tan}=-2$
Work Step by Step
a)
Function $y=f(x)=x^{2}-1$
$x_{0}=-1$
$y=f(x_{1})=x_{1}^2-1$
$y=f(x_{0})=x_{0}^2-1$
Slope of tangent line =$m_{tan}=\lim\limits_{x_{1} \to x_{0}}\frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}$
$=\lim\limits_{x_{1} \to x_{0}}\frac{(x_{1}^2-1)-(x_{0}^2-1)}{x_{1}-x_{0}}$
$=\lim\limits_{x_{1} \to x_{0}}\frac{x_{1}^2-1-x_{0}^2+1}{x_{1}-x_{0}}$
$=\lim\limits_{x_{1} \to x_{0}}\frac{x_{1}^2-x_{0}^2}{x_{1}-x_{0}}$
$=\lim\limits_{x_{1} \to x_{0}}\frac{(x_{1}+x_{0})(x_{1}-x_{0})}{x_{1}-x_{0}}$
$=\lim\limits_{x_{1} \to x_{0}}(x_{1}+x_{0})$
Applying Limit
$=(x_{0}+x_{0})$
General formula for Slope of tangent line =$m_{tan}=2x_{0}$
b)
putting the given value of $x_{0}=-1$ in the formula $m_{tan}=2x_{0}$
from ($a$)
Slope of tangent line =$m_{tan}=2(-1)$
Slope of tangent line =$m_{tan}=-2$
