Answer
$g'(1)=4$
The equation of the tangent line is $$(l):y=4x-5$$
Work Step by Step
$$g(x)=x^4-2$$
1) The derivative of $g(x)$ at a number $a$ is calculated as follows: $$g'(a)=\lim\limits_{x\to a}\frac{g(x)-g(a)}{x-a}$$
$$g'(a)=\lim\limits_{x\to a}\frac{(x^4-2)-(a^4-2)}{x-a}$$
$$g'(a)=\lim\limits_{x\to a}\frac{x^4-a^4}{x-a}$$
$$g'(a)=\lim\limits_{x\to a}\frac{(x^2-a^2)(x^2+a^2)}{x-a}$$
$$g'(a)=\lim\limits_{x\to a}\frac{(x-a)(x+a)(x^2+a^2)}{x-a}$$
$$g'(a)=\lim\limits_{x\to a}(x+a)(x^2+a^2)$$
$$g'(a)=(a+a)(a^2+a^2)$$
$$g'(a)=2a\times2a^2=4a^3$$
So $g'(1)=4\times1^3=4$
2) The equation of the tangent line $l$ of the curve $y$ at point $(1,-1)$ would have the following form $$(l):y=g'(1)x+b$$$$(l):y=4x+b$$
Since point $(1-1)$ lies in the tangent line $l$, we can find $b$ by applying the point into the equation of the tangent line $l$: $$4\times1+b=-1$$$$4+b=-1$$$$b=-5$$
Overall, the equation of tangent line $l$ is $$(l):y=4x-5$$
