If f(x) = (1 + ln x)^2 , then f '(x) =?

Question

Is this right?
(2/x)(x + 2)(x + 2 ln x)

Answer 1

Assuming f'(x) is the derivative function this is a classic case for the appliction of the chain rule dy/dx = dy/du (du/dx)

u = 1 + ln(x)

dy/du = 2 u ..... since the power rule is f'(x)^n = (n+1) x ^(n-1) right?

then f'(x) = 2 u (1+ lnx)'

du/dx= (1+ln(x))' = 0 + 1/x right?

f'(x) = dy/dx = 2( 1 + ln(x)) (1/x) .... substituting x here for u

f'(x) = (2/x) (ln(x)+1)

Answer 2

No not right .

f'(x) = 2*(1+ln x) * (1+ln x)'

(1+ln x )' = 0 + 1/x

thus
f(x)' = 2*x*(1+ln x)

Answer 3

No, but you're doing parts of it right. Use the chain rule:
(f(g(x)))' = f'(g(x))(g'(x))
So, x^2 is f, and 1+ln(x) is g.
f' = 2x
g'= 1/x
Substitute g for x in f', and write the solution:
(f(g(x)))' = 2(1+ln(x))*(1/x), which is your derivative.

Answer 4

f(x) = (1 + ln x)^2
f '(x) = 2(1 + ln x) * (0 + 1/x)
f '(x) = 2(1 + ln x)/x

Answer 5

Yours is wrong, tornado is right and
anonymous is right with theory also.

Answer 6

f'(X)=2 (1+lnX) (1/X)
your's is wrong

Answer 7

π