HELP DERIVATIVES!!!!!!!!!!!!!
if h(x)=f[f(x)] use the tables for f and f' to find the value of h'(1)
X f(x) f'(x)
1 3 2
2 1 5
3 6 7
h'(1) = f'(f(1)) f'(1) = f'(3)f'(1) = 7*2 = 14
f(x)=3,1,6
f'(x)=2,5,7
Just in case u cant read the chart
thank you!!!
To find the derivative of h(x) = f[f(x)], we need to use the chain rule.
The chain rule states that if we have a composition of two functions, say f(g(x)), then the derivative of this composition is given by the product of the derivative of the outer function f'(g(x)) multiplied by the derivative of the inner function g'(x).
In this case, the outer function f is f(x), and the inner function g is f(x) again. So, we have h(x) = f[f(x)].
Let's find h'(x):
First, we find the derivative of the outer function:
f'(f(x)) * f'(x) = f'[f(x)] * f'(x)
To find h'(1), we need to substitute x = 1 into the expression we derived above. So, we have:
h'(1) = f'[f(1)] * f'(1)
Now, let's substitute the values from the table for f and f':
f(1) = 3, f'(1) = 2
Therefore, h'(1) = f'[3] * 2
From the table, we don't have the value of f'(3), but we do have f'(2) = 5. Since f(1) = 3, and f(2) = 1, we can interpolate to find the value of f'(3) using the two given values of f'(x):
f'(3) = (f'(2) - f'(1)) / (f(2) - f(1)) * (3 - 1) + f'(1)
= (5 - 2) / (1 - 3) * (3 - 1) + 2
= 3/(-2) * 2 + 2
= -3 + 2
= -1
Now, we can substitute f'(3) = -1 into the expression for h'(1) we derived earlier:
h'(1) = f' [f(1)] * 2
= f'[3] * 2
= -1 * 2
= -2
So, the value of h'(1) is -2.