Suppose that f and g are functions that are differentiable at x = 1 and that f(1) = 2, f '(1) = −1, g(1) = −2, and g'(1) = 3. Find h'(1). h(x) = (x2 + 7)g(x)

Question

Suppose that f and g are functions that are differentiable at x = 1 and that f(1) = 2, f '(1) = −1, g(1) = −2, and g'(1) = 3. Find h'(1).

h(x) = (x² + 7)g(x)

h'(1)=???

Answer 1

h(x) = (x²+7)g(x)

By the Product Rule, h'(x) = 2xg(x) + (x²+7)g'(x).

So, h'(1) = 2g(1) + 8g'(1) = 2(-2) + 8(3) = 20