Q. 15 - Calculus

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  $f\left(x\right)=h\left(g\left(x\right)j\left(x\right)\right)$, find $f\text{'}\left(0\right)$

Step-by-Step Solution

Answer

The value of $f\text{'}\left(0\right)=-12$

Step 1. Given information:

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

Given table:

Step 2. Find f ' ( 0 ) using chain rule:

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

Hence, according to the chain rule of derivative:

$f\text{'}\left(x\right)=h\text{'}\left(g\left(x\right)j\left(x\right)\right)\times \frac{d}{dx}\left(g\left(x\right)j\left(x\right)\right)$

Apply product rule:

$$\begin{gathered} f\text{'}\left(x\right)=h\text{'}\left(g\left(x\right)j\left(x\right)\right)\times \left(g\left(x\right)j\text{'}\left(x\right)+j\left(x\right)g\text{'}\left(x\right)\right) \\ f\text{'}\left(0\right)=h\text{'}\left(g\right(0\left)j\right(0\left)\right)\times \left(g\left(0\right)j\text{'}\left(0\right)+j\left(0\right)g\text{'}\left(0\right)\right) \end{gathered}$$

From the given table we can see that

$$\begin{gathered} g\left(0\right)=2 \\ g\text{'}\left(0\right)=-2 \\ j\left(0\right)=0 \\ j\text{'}\left(0\right)=-2 \end{gathered}$$

Substitute all these values in the above derivative:

$$\begin{gathered} f\text{'}\left(0\right)=h\text{'}(2\times 0)\times \left(2(-2)+0(-2)\right) \\ =h\text{'}\left(0\right)\left(-4+0\right) \\ =-4\times h\text{'}\left(0\right) \end{gathered}$$

In table $h\text{'}\left(0\right)=3$

So, $$\begin{gathered} f\text{'}\left(0\right)=-4\times 3 \\ =-12 \end{gathered}$$