f(0)=4 , f′(0)=9 , f(1)=4 , f′(1)=5 , g(1)=3 , g′(1)=−3 H(x)=f(ln(x))+ln(g(x)) find the derivative at x=1 H'(1)=???

Answer 1

This question is testing your knowledge of the chain rule.

H(x) = f(ln(x)) + ln(g(x))

Here, you can take the derivative of both sides and apply the chain rule

H'(x) = f'(ln(x))* (ln(x))' + 1/g(x) * g'(x)

H'(x) = f'(ln(x))* (1/x) + 1/g(x) * g'(x)

H'(1) = f'(ln(1))* (1/1) + 1/g(1) * g'(1)

H'(1) = f'(0) + 1/g(1) * g'(1)

H'(1) = 9 + 1/3 * (-3)

H'(1) = 8