Question

Suppose g, h, and j are differentiable functions with the values for the function and derivative given in the following table:

Use the table to calculate the values of the derivatives listed in Exercises 9–16.

If $f\left(x\right)=g(x^3-6)$, find$f\text{'}\left(2\right)$.

Short answer

The value of$f\text{'}\left(2\right)=-12$

Step-by-step solution

Step 1. Given information: 

Function is:$f\left(x\right)=g\left(x^3-6\right)$

Given table:

Step 2. Find f'(2) using chain rule:   

Since$f\left(x\right)=g\left(x^3-6\right)$

Hence, according to the chain rule of derivative:

$$\begin{gathered} f\text{'}\left(x\right)=g\text{'}(x^3-6)\times \frac{d}{dx}\left(x^3-6\right) \\ f\text{'}\left(x\right)=g\text{'}\left(x^3-6\right)(3x^2-0) \\ f\text{'}\left(2\right)=g\text{'}(2^3-6)(3\times 2^2) \\ f\text{'}\left(2\right)=g\text{'}(8-6)\times 12 \\ f\text{'}\left(2\right)=12g\text{'}\left(2\right) \end{gathered}$$

From the given table we can see that

$g\text{'}\left(2\right)=-1$

Substitute all these values in the above derivative:

$$\begin{gathered} f\text{'}\left(2\right)=12(-1) \\ =-12 \end{gathered}$$