Answer:
h'(1) = 19
Explanation:
if h(x) = f(x) + g(x), and f(x) = x³ and g(x) = 4x⁴,
then h(x) = (x³) + (4x⁴)
h(x) = x³ + 4x⁴
h'(x) is the derivative of h(x), which = 3x² + 16x³. Thus, h'(1) = 3(1)² + 16(1)³ = 19
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