Possible mistake during differentiation? Please check

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    Differentiation Mistake
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Discussion Overview

The discussion revolves around the differentiation of the function f(x) = ln(x)/√x, specifically focusing on the calculation of its first derivative f'(x). Participants are examining the application of the quotient rule versus the product rule and identifying potential mistakes in the differentiation process.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant claims to have used the quotient rule to find f'(x) but received a different result from Wolfram Alpha, which used the product rule.
  • Another participant suggests that the original poster did not differentiate √x correctly and points out that the quotient rule requires dividing by the square of the denominator.
  • Concerns are raised about the numerator in the derivative calculation, specifically questioning how the term -√x/2 was derived.
  • There is a discussion about the correct application of the power rule for differentiating √x, with one participant explaining their reasoning for arriving at -√x/2.
  • A later reply acknowledges a transcription error in the original post regarding the denominator of the derivative.

Areas of Agreement / Disagreement

Participants express differing views on the correct application of differentiation rules, with no consensus reached on the original poster's method or the correctness of their derivative calculation.

Contextual Notes

There are unresolved issues regarding the proper application of the quotient rule and the differentiation of √x, as well as the implications of transcription errors in mathematical expressions.

Who May Find This Useful

This discussion may be useful for students or individuals learning about differentiation techniques, particularly in the context of applying the quotient and product rules.

Dominathan
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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!
 
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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.
 
Dominathan said:
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] ?
 
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
 
gb7nash said:
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.
 
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