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Logarithm differentiation + chain rule

  1. Aug 26, 2014 #1
    For this function

    [itex]y=\sqrt{2ln(x)+1}[/itex]

    if I use the chain rule properly, should I be getting this answer?

    [itex]\frac{dy}{dx}=\frac{2}{x} \times \frac{1}{2} \times \frac{1}{\sqrt{2ln(x)+1}}[/itex]

    My aim of doing this is to verify that

    [itex]\frac{dy}{dx}=\frac{1}{xy}[/itex]
     
  2. jcsd
  3. Aug 27, 2014 #2

    ShayanJ

    User Avatar
    Gold Member

    That's right and equal to what you want!
     
  4. Aug 31, 2014 #3

    Exactly what Shaun said.
    If you equate that to what information you need.

    BINGO! You will get your answer
     
  5. Aug 31, 2014 #4

    Mark44

    Staff: Mentor

    As a simpler alternative to using the chain rule, you can do this:

    ##y = \sqrt{2\ln(x) + 1} ##
    ##\Rightarrow y^2 = 2\ln(x) + 1##
    Differentiating implicitly,
    ##2y\frac{dy}{dx} = \frac{2}{x}##
    ##\Rightarrow \frac{dy}{dx} = \frac{1}{xy}##
     
  6. Sep 1, 2014 #5
    Yes, that is correct. Just to show the math:

    d/dx[√2ln(x)+1]
    =1/2√2ln(x)+1 * 2/x
    =2/x * 1/2 * 1/√2ln(x)+1
     
  7. Sep 1, 2014 #6

    Mark44

    Staff: Mentor

    You are missing several sets of parentheses.
    Most would interpret the part in brackets above as √2 * ln(x)+1, rather than √(2ln(x)+1).
    Most would interpret the above as
    1/2 * √2ln(x) + 2/x, which I don't think is what you intended.
     
  8. Sep 1, 2014 #7
    You're right, perhaps I should have put 2ln(x) + 1 in parentheses, but I think he gets what I mean haha. Nonetheless, I appreciate the correction. Thank you!
     
  9. Sep 2, 2014 #8
    Hm, how does this take in to account the fact that squaring the equation will give multiple values?

    ##y^2=2\ln(x)+1## is a set of two equations, namely $$y= \sqrt{2\ln(x) + 1}$$ and $$y= -\sqrt{2\ln(x) + 1}$$
    Will the derivative also be multiple valued? It doesn't seem like it since ##\frac{1}{xy}## is a singled valued funtion. What am I missing?
     
  10. Sep 3, 2014 #9

    Mark44

    Staff: Mentor

    Actually it doesn't in this case. That's something that I considered but didn't mention before. In the first equation, the square root term is necessarily nonnegative, which means that y must also be nonnegative. So squaring both sides doesn't introduce extraneous solutions that weren't present in the original equation.
    I don't understand what you're asking. The derivative will not be multiple valued.
     
  11. Sep 4, 2014 #10
    What I'm trying to ask is that since we found the derivative is ##\frac{dy}{dx}=\frac{1}{xy}##, we have two options for ##y## now, the positive or negative root, since we squared the term to make the differentiation simpler. Why do we necessarily choose the positive one? I know I'm just being technical, sorry about my arrogance, but it's a pet-peeve of mine.
     
  12. Sep 5, 2014 #11

    Mark44

    Staff: Mentor

    No we don't in this case, which I explained in post #9. Since y is the square root of some expression, y is necessarily nonnegative. Take a look again at what I wrote, and see if that answers your questions.


     
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