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 $f\left(x\right)=j\left(2x\right)$, find$f\text{'}(-1)$

Short answer

The value of$f\text{'}(-1)=0$

Step-by-step solution

Step 1. Given information:  

Function is: $f\left(x\right)=j\left(2x\right)$

Given table:

Step 2. Find f'(-1) using chain rule:  

Since $f\left(x\right)=j\left(2x\right)$

Hence, according to the chain rule of derivative:

$$\begin{gathered} f\text{'}\left(x\right)=j\text{'}\left(2x\right)\times \frac{d}{dx}\left(2x\right) \\ f\text{'}\left(x\right)=j\text{'}\left(2x\right)\times 2 \\ f\text{'}(-1)=j\text{'}(-2)\times 2 \end{gathered}$$

From the given table we can see that

$j\text{'}(-2)=0$

Substitute all these values in the above derivative:

$$\begin{gathered} f\text{'}(-1)=0\times 2 \\ =0 \end{gathered}$$