If f(x)=ln(x^3+1) find f′(7) correct to at least five decimal places.

Answer 1

f(x) = ln(x^3 +1)

f'(x) = (1/(x^3+1))(3x^f2) = 3x^2/(x^3+1)

f'(7) = 3(49)/(343+1) = 147/344 = about .4273255814...

f'(7) = about 0.42733

use the chain rule. good on tires in heavy snow too.

f(u) = lnu

f'(x) = (1/u)(du/dx)= u'(x)/u

then plug in x^3+1 for u, 3x for u', and 7 for x