All,

This question is from "Forgotten Calculus"--it's on page 310.

If f(x)= x lnx-x, find f'(x).  Hint: The first term is a product.

Okay, so I factored out x, and was left with:

f(x) = x(lnx-1)

Using the product rule:

If f(x)=F(x)S(x) then f'(x)=F(x)S'(x) + S(x)F'(x)

I decided that F(x) was "x" and F'(x) is 0, also, S(x) is lnx, so
S'(x) would be 1/x

This gives me:

f'(x) = x(1/x) + lnx(0) 

Which works out to f'(x)= x/x or 1.

The book states the answer is f'(x)= lnx

Where did I make my mistake?

Regards,

Herkdrvr

Hi herkdrvr!!


It is all ok until the use of the product rule. This rules states that:

If f(x) = F(x)*S(x) then f'(x) = F(x)*S'(x) + S(x)*F'(x)

You decided that:

F(x) = x ==> F'(x) = 1  (mistake here: (x)' = 1 not zero)
and
S(x) = ln(x)-1 ==> S'(x) = (ln(x)- 1)' = 
                         = [ln(x)]'- [1]' =
                         = 1/x - 0 =
                         = 1/x
(mistake here: S(x) = ln(x)-1 , not ln(x) alone)

Then we have:
f'(x) = F(x)*S'(x) + S(x)*F'(x) =
      = x*1/x + (ln(x)-1)*1 =
      = 1 + ln(x) - 1 =
      = ln(x)


I hope that this helps you. Feel free to request for a clarification
if you need it.

Regards.
livioflores-ga

Prompt reply, and clearly explained!  Thank you!!!

Herkdrvr

If F(x) = x, then F'(x) cannot be zero.

Keep in mind you had written this product:

f(x) = x * (ln(x) - 1)

So, while the first factor is certainly x, the second factor is not ln(x).

-- mathtalk-ga