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

h(x) = f(lnx) + ln(g(x))

h'(x) = f'(lnx)(lnx)' + g'(x) / g(x) = f'(lnx) / x + g'(x) / g(x)

So, h'(1) = f'(ln1) / 1 + g'(1) / g(1) = f'(0) + (-3) / 3 = 9 - 1 = 8

Answer 2

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

H'(1) = f'(ln1 + ln(g(1)))·(1/1 + g'(1)/g(1)) = f'(ln3)(1 - 3/3) = f'(ln3)·0 = 0