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 + 12)g(x)

Answer 1

h(x) = (x² + 12)g(x)

h'(x) = 2x(g(x)) + (x² + 12)g'(x)

h'(x) = 2(-2) + (1 + 12)(3)

h'(x) = -4 + 39

h'(x) = 35