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

Question

possible solutions...

(2/x)(x + 1)(x + ln x)

(2/x)(1 + ln x)

(2/x)(x + 2)(x + 2 ln x)

(2/x)(x + 2 ln x)

(4/x)(1 + 2 ln x)

(4/x)(x + 2)(x + 2 ln x)

Answer 1

f(x) = (x + ln(x))^2

We have the squared function to deal with, and then the chain rule for the functions inside.

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

We can simplify this to

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

Answer 2

its

2(x + ln x)(1 + 1/x)

using the chain rule


differentiate the bracket, so times by the power and reduce the power 2 (x + ln x)

then times by the differential of the bracket (ie times by (1 + 1/x))

and u get

2(x + ln x)(1 + 1/x)

Answer 3

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

power rule first then take the derivative of the inside
derivative of x = 1 ; derivative of lnx = 1/x

Answer 4

mastcp is right

Answer 5

Do your own homework jackass...