If h(x) = Squareroot 7 + 6f(x), where f(1) = 7 and f'(1) = 2, find h'(1)

Answer 1

h(x) = √7 + 6f(x)
Now that seems a bit ridiculous, so I will go with
h(x) = √(7+6f(x))
by the chain rule,
h'(x) = 1/(2√(7+6f(x)) * 6f'(x)
so h'(1) = 1/(2√(7+6*7)) * 6*2 = 1/(2*7) * 12 = 6/7

Answer 2

ahh thank you!! The square root was kind of throwing me off