Possible mistake during differentiation? Please check

  • Thread starter Dominathan
  • Start date
  • #1

Main Question or Discussion Point

For a function f(x), I have to determine intervals of increase/decrease, find local max(s)/min(s), and find intervals of concavity. The first thing I'm doing in this is to write out f'(x) and f''(x).

f(x) = [itex]ln(x)/\sqrt{x}[/itex]

For f'(x), I used the quotient rule and received f'(x) = (([itex]\frac{1}{x}[/itex][itex]\sqrt{x}[/itex])-([itex]\frac{-\sqrt{x}}{2}[/itex]ln(x))) / 2

However, I plugged f(x) into wolfram alpha and it gave me: [itex]\frac{2-ln(x)}{2x^{3/2}}[/itex]

I don't understand the difference? I thought I had done this correctly but apparently not? Wolfram alpha used the product rule. Is there some kind of algebraic gymnastics I'm forgetting about? I really want to understand where my error was made, not just which is the correct answer. Thanks!
 

Answers and Replies

  • #2
993
13
In using the quotient rule you are not differentiating [itex]\sqrt{x}[/itex] corretly and the quotient rule requires you to divide by the square of the denominator.
 
  • #3
gb7nash
Homework Helper
805
1
f(x) = [itex]ln(x)/\sqrt{x}[/itex]

For f'(x), I used the quotient rule and received f'(x) = (([itex]\frac{1}{x}[/itex][itex]\sqrt{x}[/itex])-([itex]\frac{-\sqrt{x}}{2}[/itex]ln(x))) / 2
I see a couple of issues here:

1) How did you get [itex]\frac{-\sqrt{x}}{2}[/itex] in the numerator? What is the derivative of [itex]\sqrt{x}[/itex] ?

2) How did you get 2 in the denominator? [itex](\sqrt{x})^{2} = [/itex] ?
 
  • #4
724
0
Using quotient rule, you should get [itex]\frac{\frac{\sqrt{x}}{x} - \frac{\ln{x}}{2 \sqrt{x}}}{x}[/itex] which simplifies to what you got from WA. It looks like you messed up on the derivative of [itex]\sqrt{x}[/itex] and on the bottom of the quotient rule. http://en.wikipedia.org/wiki/Quotient_rule
 
  • #5
I see a couple of issues here:

1) How did you get [itex]\frac{-\sqrt{x}}{2}[/itex] in the numerator? What is the derivative of [itex]\sqrt{x}[/itex] ?


2) How did you get 2 in the denominator? [itex](\sqrt{x})^{2} = [/itex] ?
1.
The following is my logic for the answer I received:
  1. [itex]\sqrt{x}[/itex] = x[itex]^{1/2}[/itex]
  2. Using the power rule I bring the 1/2 out as a coefficient, and subtract one from the numerator : [itex]\frac{1}{2}[/itex][itex]x^{-1/2}[/itex]
  3. I simplified to : [itex]\frac{-\sqrt{x}}{2}[/itex]

2.
My bad! I did a poor job transcribing it from my notebook to the syntax used on this site. It was (obviously) my first post, but far from my last! I meant to put "x" as the denominator, that was a mental slip.
 
  • #6
993
13
Welcome to the PhysicsForums! One may sometimes swallow some water but that is the way to learns to swim!
 

Related Threads on Possible mistake during differentiation? Please check

Replies
7
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
3
Views
457
Replies
10
Views
918
Top