Chain Rule and Implicit Differentiation

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 $x=4$?

Enter your answer in the form $\verb#y=mx+b#$.

How am I suppose to begin this question?

Answer 1

The information you are given is that the slope of the tangent line at $x=4$ is $-3$ (that is, $g'(4)=-3$), and that $g(4)=-3(4)+11=-1$. By the Chain Rule, $$ (g(x)^3)'|_{x=4}=3g(4)^2\,g'(4)=3(-1)^2(-3)=-9. $$

Answer 2

With $$y(x)=g(x)^3$$ we get the slope at $$x=4$$: $$y'(x)=3g(x)^2g'(x)$$ so $$y'(4)=3g(4)g'(4)$$ Since we have given the equation of the tangent line $$y=-3x+11$$ it must be $$g'(4)=-3$$ Can you finish?

Answer 3

The problem of finding a tangent line can always be broken down into finding the slope of the tangent line (the derivative) and the point at which you are finding the line. This problem gives the information needed in rather obscure ways so it is important to focus on exactly the information needed in order to not get lost.

We notice that the equation for the tangent of g(x) at x = 4 is given. Two pieces of information can be inferred as follows. 1. The derivative g'(4) = -3 (the slope of the tangent line. Second g(4) = -1. In order to calculate the tangent line for $g(x)^3$, we need to find the value of the derivative at x = 4 and compute the output value. $\frac{d}{dx}g(x)^3$ by the chain rule is equal to $3g(x)^2*g'(x) = 3(-1)^2*(-3) = -9$ Thus we know the slope of the tangent line is equal to -9. We know the x value to be equal to 4, thus $g(4)^3 = (-1)^3 = -1$

The line can be written as : $y = -9(x-4)-1$ or in ax+b form, y = -9x +35