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 $f\left(x\right)=h\left(g\right(j\left(x\right)\left)\right)$, find $f\text{'}\left(1\right)$.

Short answer

The value of$f\text{'}\left(1\right)=4$

Step-by-step solution

Step 1. Given information:  

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

Given table:

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

Since$f\left(x\right)=h\left(g\right(j\left(x\right)\left)\right)$

Hence, according to the chain rule of derivative:

$$\begin{gathered} f\text{'}\left(x\right)=h\text{'}\left(g\right(j\left(x\right)\left)\right)\times g\text{'}\left(j\right(x\left)\right)\times j\text{'}\left(x\right) \\ f\text{'}\left(1\right)=h\text{'}\left(g\right(j\left(1\right)\left)\right)\times g\text{'}\left(j\right(1\left)\right)\times j\text{'}\left(1\right) \end{gathered}$$

From the given table we can see that

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

Substitute all these values in the above derivative:

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

From table,

$$\begin{gathered} g(-2)=1 \\ g\text{'}(-2)=2 \end{gathered}$$

So,$$\begin{gathered} f\text{'}\left(1\right)=h\text{'}\left(1\right)\times 2\times (-1) \\ =-2h\text{'}\left(1\right) \\ =-2\times \left(-2\right) \\ =4 \end{gathered}$$