if f(x)=ln(x^3 +1) find f'(3), correct to 2 decimal places

Answer 1

Use the chain rule, with u = x^3 + 1
f(x) = ln u(x)

df/dx =df/du*du/dx
= 1/u * 3x^2
= 3x^2/[x^3 +1]

Now plug in x = 3 and turn the crank.