f(x) and g(x) are a differentiable function for all reals and h(x) = g[f(3x)]. The table below gives selected values for f(x), g(x), f '(x), and g '(x). Find the value of h '(1)

Numerical answers expected!

x 1 2 3 4 5 6
f(x) 0 3 2 1 2 0
g(x) 1 3 2 6 5 0
f'(x) 3 2 1 4 0 2
g'(x) 1 5 4 3 2 0

To find the value of h'(1), we need to use the chain rule. The chain rule states that if y = g(u) and u = f(x), then dy/dx = dy/du * du/dx.

In this case, h(x) = g[f(3x)], so we can think of h(x) as g(u), where u = f(3x).

To find h'(x), we need to find h'(x) = dy/dx.

First, let's find du/dx. We know that u = f(3x), so du/dx = d(f(3x))/dx.

Using the chain rule, we have du/dx = f'(3x) * d(3x)/dx.

d(3x)/dx = 3 since the derivative of 3x with respect to x is 3.

Therefore, du/dx = f'(3x) * 3.

Now, let's find dy/du. We have y = g(u), so dy/du = g'(u).

To find h'(x) = dy/dx, we use the chain rule: h'(x) = dy/dx = dy/du * du/dx = g'(u) * f'(3x) * 3.

To find h'(1), we substitute x = 1 into the expression for h'(x): h'(1) = g'(u) * f'(3).

Looking at the table, when x = 1, we have u = f(3), which is f(3 * 1) = f(3) = 2.

Therefore, h'(1) = g'(2) * f'(3) * 3.

From the table, we see that when u = 2, g'(u) = g'(2) = 4.

Similarly, when x = 3, f'(3) = 1.

Substituting these values, we have h'(1) = 4 * 1 * 3 = 12.

Therefore, the value of h'(1) is 12.