Question

Suppose g(x), h(x), and j(x) are differentiable functions with the values of the function and its derivative given in the following table:

x	g(x)	h(x)	j(x)	g'(x)	h'(x)	f'(x)
-1	3	0	1	-1	-2	-2
0	2	3	0	-2	3	-2
1	0	-1	-2	-2	-2	-1
2	-2	-2	-3	-1	0	2
3	-3	0	1	0	1	2

f(x)=g(x)(h(x)+j(x)) find f'(2)

Short answer

The value is$f\text{'}\left(2\right)=1$

Step-by-step solution

Given information

We are given a table

x	g(x)	h(x)	j(x)	g'(x)	h'(x)	f'(x)
-1	3	0	1	-1	-2	-2
0	2	3	0	-2	3	-2
1	0	-1	-2	-2	-2	-1
2	-2	-2	-3	-1	0	2
3	-3	0	1	0	1	2

Calculate

$$\begin{gathered} f\text{'}\left(x\right)=g\text{'}\left(x\right)\left[h\right(x)+j(x\left)\right]+g\left(x\right)[h\text{'}(x)+j(x\left)\right] \\ f\text{'}\left(2\right)=g\text{'}\left(2\right)\left[h\right(2)+j(2\left)\right]+g\left(2\right)[h\text{'}(2)+j(2\left)\right] \\ f\text{'}\left(2\right)=-1[-5]+(-2)[0+2] \\ f\text{'}\left(2\right)=5-4 \\ f\text{'}\left(2\right)=1 \end{gathered}$$