Question

let r(x) = f(g(h(x))), where h(1) = 2, g(2) = 4, h'(1) = 4, g'(2) = 4, and "f '(4) = 7". find r'(1). r'(1)

Answer 1

Answer: To find r'(1), we need to use the chain rule of differentiation, which states that:

r'(x) = f'(g(h(x))) * g'(h(x)) * h'(x)

In this case, we know that:

h(1) = 2, g(2) = 4, h'(1) = 4, g'(2) = 4, and f '(4) = 7

Therefore:

r'(1) = f'(g(h(1))) * g'(h(1)) * h'(1) = f'(g(2)) * g'(2) * h'(1) = 7 * 4 * 4 = 112

So, the derivative of r(x) with respect to x at x = 1 is 112

Explanation: