Let f and g be differentiable functions such that
f(1) = 4, g(1) = 3, f '(3) = -5
f '(1) = -4, g '(1) = -3, g '(3) = 2
If h(x) = f(g(x)), then h '(1) = ?
To find h'(1), we need to use the chain rule, which states that if h(x) = f(g(x)), then h'(x) = f'(g(x)) * g'(x).
First, let's find g'(3). We are given g'(1) = -3 and g'(3) = 2. Since g(x) is differentiable, we can use the Mean Value Theorem to find g'(3) by considering the interval [1, 3].
The Mean Value Theorem states that if a function is continuous on the closed interval [a, b], and differentiable on the open interval (a, b), then there exists at least one number c in the interval (a, b) such that f'(c) = [f(b) - f(a)] / (b - a).
Applying the Mean Value Theorem to g(x) on the interval [1, 3], we have:
(g(3) - g(1)) / (3 - 1) = g'(c)
(3 - 1) / (3 - 1) = g'(c)
2 = g'(c)
So, g'(3) = 2.
Now, let's find h'(1) using the chain rule:
h'(1) = f'(g(1)) * g'(1)
Substituting the given values, we have:
h'(1) = f'(3) * g'(1)
We are given f'(1) = -4 and f'(3) = -5.
So, h'(1) = -5 * (-3) = 15.
Therefore, h'(1) = 15.