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Need help with combination of product and chain rule

  1. Sep 13, 2006 #1
    I need help differentiating [tex] y=\sqrt{x-2}\sqrt{x+1} [/tex]

    I am using a mix of the chain rule and product rule which my textbook for school wants me to use for this. So suggestions for differant ways of approaching it won't help :P Anyways, thanks in advance for looking over it. It's my first attempt at posting in latex so lets hope it works!

    For my work I have:



    And that is exactly what they have in my textbook, so thats good. But...

    For the third part my textbook has


    This is the step that I am not understanding, it's simple algebra but I am still not quite getting it :( Can someone help?

    Then the last step is

    =[tex] \frac{2x-1}{2(x+1)^{1/2}(x-2)^{1/2}}[/tex] Which I somewhat understand, but not really.

    PS - For the third step I am getting

    =[tex](x-2)^{1/2}(\frac{1}{2})(x+1)^{-1/2}+(\frac{1}{2})(x+1)^{1/2}(x-2)^{-1/2} [/tex]
    Last edited: Sep 13, 2006
  2. jcsd
  3. Sep 13, 2006 #2
    For the third part, you're multiplying and dividing by [tex](x+1)^{-1/2}(x-2)^{\frac{-1}{2}}[/tex]. Try it. The fourth step is just writing the previous expression in a neat way (Remember, [tex]x^{-n} = \frac{1}{x^n}[/tex]).

    Btw, is it 3 or 2 that's in the first term? ;)
  4. Sep 13, 2006 #3
    Fixed :). Now I gotta sit and think for a bit.
  5. Sep 13, 2006 #4
    Aha! Finally I get it, simply set them up as 1/whatever^(1/2) then cross multiply to add :P Don't know why it didn't sink in immediately after I read your post but sometimes math is like that I guess. Thankyou Neutrino.

    Anyways I have one more question for the day..

    The Question is:

    Differentiate y=[tex] \frac{3-x}{\sqrt{x^{2}-2x}} [/tex]

    So here I am practicing the quotient rule instead of the product rule... I could of course use the product rule on y=(3-x)(x^2-2x)^(-1) but that would mean that I don't learn how to use the quotient rule. :P

    The quotient rule is, of course [tex] \frac {v\frac {du}{dx} - u \frac {dv}{dx}}{v^{2}} [/tex]

    Anyways, (this is an example problem by the way, not a homework question) For their first step they have:

    (dy)/(dx)=[tex] \frac {\sqrt{x^{2}-2x} \frac {d}{dx}(3-x) - (3-x) \frac {d}{dx} (\sqrt{x^{2}-2x}}{x^{2}-2x} [/tex]

    I understand this. The derivative of y with respect to x is [tex] \sqrt{x^{2}-2x} [/tex] multiplied by the derivative of (3-x) (which is of course -1) minus (3-x) multiplied by the derivative of [tex] \sqrt{x^{2}-2x} [/tex] (which is (2x-2)) But then for their second step they have:

    [tex] \frac {(x^{2}-2x)^{1/2}(-1)-(3-x)(\frac{1}{2})(x^{2}-2x)^{-1/2}(2x-2)}{(x^2-2x)} [/tex]

    I understand the first part on the numerator, (x^2-2x)^(1/2)(-1) since that is exactly what I stated should happen, the x^2-2x term is being multiplied by the derivitive of (3-x). But then the second part I don't quite understand... Why is (3-x) being multiplied by the derivative AND another form of the derivative? The power rule states that (x^2-2x)^(1/2) has a derivative of (1/2)(x^2-2x)^(-1/2) or the nx^(n-1)... I think this is being caused by me trying to grasp all the rules of the derivative in one day, right now it's like a mish mash of semi-remembered ideas. lol.
    Last edited: Sep 13, 2006
  6. Sep 13, 2006 #5
    The chain rule.

    Suppose [tex]t = \sqrt{x^2-2x}[/tex] and [tex]u = x^2 - 2x[/tex].
    Therefore, [tex]t = u^\frac{1}{2}[/tex]. Now, use the chain rule to differentiate t wrt x.
  7. Sep 13, 2006 #6


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    Dearly Missed

    This is perhaps best handled by the logarithm rule.
  8. Sep 13, 2006 #7

    Thanks alot guys. Neutrino, much appreciated. I woke up today and read this thread and my mistake is much more obvious now. For some reason I was thinking that I was only using the power rule but of course the power rule is only for equations such as y=x^2 whereas the chain rule is for equations such as [tex] \sqrt {x-a} [/tex]. Anyways, thanks!!
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