Answer
$g'(1) = 4$
Equation of the tangent line to the curve $y = x^4 - 2$ at the point $(1, -1): y=4x-5$
Work Step by Step
Given $g(x)=x^4-2$
\begin{gather*}
g'(1) = \lim\limits_{x \to 1} \dfrac{g(x)-g(1)}{x-1} = \lim\limits_{x \to 1} \dfrac{x^4-2 - (1^4-2)}{x-1} = \lim\limits_{x \to 1} \dfrac{x^4-2 - (-1}{x-1} = \lim\limits_{x \to 1} \dfrac{x^4-2 +1}{x-1} = \lim\limits_{x \to 1} \dfrac{x^4-1}{x-1} = \lim\limits_{x \to 1} \dfrac{(x^2-1)(x^2+1)}{x-1} = \lim\limits_{x \to 1} \dfrac{(x-1)(x+1)(x^2+1)}{x-1} = \lim\limits_{x \to 1} [(x+1)(x^2+1)] = (1+1)(1^2+1) = \\ (2)(2) = 4 \longrightarrow g'(1) = 4
\end{gather*}
From the point-slope equation, we have:
\begin{gather*}
y-g(1)=g'(1)(x-1) \longrightarrow y-(1^4-2)=4(x-1) \longrightarrow y+1=4x-4 \longrightarrow y= 4x-5
\end{gather*}
