Question: The tangent line to the graph of \(y = g(x)\) at \(x=4\) has equation \(y=-3x+11\). What is the equation of the tangent line to the graph of \(y = g(x)^3\) at \(x = 4\)?
I know that the slope is the derivative of \(g(x)\). I'm treating \(g(x)^3\) as a variable (\(g(x)^3 = x^3\)); therefore, the derivative/slope of that would be \(3x^2\), but I'm not sure where to go from there. I have an idea to find \(g(4)\) using the equation first, but I'd still appreciate some help. :)
+118687 y=g(x)
y'=g'(x)
g'(4)=-3
The tangent of g(x) is y=-3x-11 and when x=4 y=-1 so (4,-1) is on the tangent point.
When x=4
g(4)=-1
[g(4)]^3 = (-1)^3 = -1
\(\frac{d}{dx}[g(x)]^3=3[g(x)]^2*g'(x)\\ \frac{d}{dx}[g(4)]^3=3[g(4)]^2*g'(4)\\ \frac{d}{dx}[g(4)]^3=3[-1]^2*-3\\ \frac{d}{dx}[g(4)]^3=3*1*-3\\ \frac{d}{dx}[g(4)]^3=-9\\ \)
so the gradient of the tangent of [g(x)]^3 at x=4 is -9
I did do a sample graph to check it worked at least for that example. It did 
https://www.geogebra.org/classic/u7dfheht
LaTex:
\frac{d}{dx}[g(x)]^3=3[g(x)]^2*g'(x)\\
\frac{d}{dx}[g(4)]^3=3[g(4)]^2*g'(4)\\
\frac{d}{dx}[g(4)]^3=3[-1]^2*-3\\
\frac{d}{dx}[g(4)]^3=3*1*-3\\
\frac{d}{dx}[g(4)]^3=-9\\
