A table of values for f,g,f′, and g′ are given in the table below:

x f(x) g(x) f′(x) g′(x)
5 0 -4 -9 5
0 5 0 -6 9
-4 5 5 9 -5
If h(x)=f(g(x)), find h′(5),
If H(x)=g(f(x)), find H′(0)

To find h′(5), we need to find the derivative of h(x) = f(g(x)) with respect to x and then evaluate it at x = 5.

Given:
f(x) table:
x f(x)
5 0
0 5
-4 5

g(x) table:
x g(x)
5 -4
0 0
-4 5

f′(x) table:
x f′(x)
5 -9
0 -6
-4 9

g′(x) table:
x g′(x)
5 5
0 9
-4 -5

To find h′(x), we'll use the chain rule:

h(x) = f(g(x))

h′(x) = f′(g(x)) * g′(x)

Substituting from the tables:

h′(5) = f′(g(5)) * g′(5)
= f′(-4) * 5
= -9 * 5
= -45

Therefore, h′(5) = -45.

Now, let's find H′(0).

H(x) = g(f(x))

H′(x) = g′(f(x)) * f′(x)

Substituting from the tables:

H′(0) = g′(f(0)) * f′(0)
= g′(5) * (-6)
= 5 * (-6)
= -30

Therefore, H′(0) = -30.

To find h(x) and H(x), we need to use the given information about the functions f(x) and g(x) and their derivatives. Let's start by understanding what h(x) and H(x) represent.

h(x) = f(g(x)) means that we substitute the value of g(x) into the function f(x). So, h(x) is a composition of the functions f(x) and g(x).

H(x) = g(f(x)) means that we substitute the value of f(x) into the function g(x). So, H(x) is also a composition of the functions g(x) and f(x).

Now, let's find h′(5) and H′(0) using the given table of values for f(x), g(x), f′(x), and g′(x).

For h′(5), we need to find the derivative of h(x) with respect to x and then evaluate it at x = 5. Since h(x) is a composition of functions, we can use the chain rule to differentiate it.

h′(x) = f′(g(x)) * g′(x)

Using the values from the table, we have:

h′(5) = f′(g(5)) * g′(5)

Using the table values, g(5) = -4. So, we substitute g(5) into f′(x) and g′(x) terms:

h′(5) = f′(-4) * g′(5)

From the table, f′(-4) = -9 and g′(5) = 5. So we substitute these values:

h′(5) = -9 * 5 = -45

Therefore, h′(5) = -45.

Now, let's find H′(0) using a similar approach.

H′(x) = g′(f(x)) * f′(x)

Using the values from the table, we have:

H′(0) = g′(f(0)) * f′(0)

Using the table values, f(0) = 5. So, we substitute f(0) into g′(x) and f′(x) terms:

H′(0) = g′(5) * f′(0)

From the table, g′(5) = 5 and f′(0) = -6. So we substitute these values:

H′(0) = 5 * -6 = -30

Therefore, H′(0) = -30.