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B (ASK) Implicit Differentiation

  1. Aug 17, 2016 #1
    What is ##\frac{d}{dx}(\frac{x}{y^2})##?

    Please tell me is it correct or not:

    ##\frac{d}{dx}(\frac{x}{y^2}) = \frac{[\frac{d}{dx}(x)] ⋅ (y^2) - (x) ⋅ [\frac{d}{dx} (y^2)]}{(y^2)^2}##
    ## = \frac{(x) ⋅ (y^2) - (x) ⋅ (\frac{d}{dy} (y^2)) ⋅ \frac{dy}{dx}}{y^4}##
    ##= \frac{xy^2 - (x)(2y)(\frac{dy}{dx})}{y^4}##
    ##= \frac{xy^2 - 2xy(\frac{dy}{dx})}{y^4}##
     
  2. jcsd
  3. Aug 17, 2016 #2

    fresh_42

    Staff: Mentor

    Do you know what ##\frac{d}{dx}x## is? And isn't there another ##y## that could be canceled?
     
  4. Aug 17, 2016 #3
    I think you made a typo - what is ## \frac{d}{dx}(x) ## ?

    P.S - fresh_42 posted at the time I was typing. Yes a y can be cancelled.
     
  5. Aug 17, 2016 #4
    Yes, sorry, I've made a typo:

    ##\frac{d}{dx}(\frac{x}{y^2}) = \frac{[\frac{d}{dx}(x)] ⋅ (y^2) - (x) ⋅ [\frac{d}{dx} (y^2)]}{(y^2)^2}##
    ##= \frac{(1) ⋅ (y^2) - (x) ⋅ (\frac{d}{dy} (y^2)) ⋅ \frac{dy}{dx}}{y^4}##
    ##= \frac{y^2 - (x)(2y)(\frac{dy}{dx})}{y^4}##
    ##= \frac{y^2 - 2xy(\frac{dy}{dx})}{y^4}##


    What y could be cancelled?
     
  6. Aug 17, 2016 #5

    fresh_42

    Staff: Mentor

    You differentiated something with ##y^{-2}## which would give you something with ##y^{-3}## but your result is something with ##y^{-4}##.
     
  7. Aug 17, 2016 #6
    Do you mean the ##y^4## in the denominator?

    What's wrong with that?

    Isn't I am already correct?

    The quotient rule is ##\frac{u'v - uv'}{v^2}## where ##v = y^2##, am I right?
     
  8. Aug 17, 2016 #7

    fresh_42

    Staff: Mentor

    Nothing. If you like, you may even write ##y^{100}##. Just said it's not the usual way. And you don't have to shout here.
     
  9. Aug 17, 2016 #8
    I am not shouting. So, what do you mean with another ##y## that could be cancelled?

    Please explain, I don't understand.
     
  10. Aug 17, 2016 #9

    fresh_42

    Staff: Mentor

    You have a common factor ##y## in each summand of the nominator and four of them in the denominator.
    Or if you like to proceed with formulas: ##ab+ac=a(b+c)## and ##\frac{y(y-2xy')}{y^4}=\frac{y-2xy'}{y^3}##.
     
  11. Aug 17, 2016 #10
    OK, but is it correct of what I've done so far?
     
  12. Aug 17, 2016 #11

    fresh_42

    Staff: Mentor

    Yes.
    And in case you want to differentiate it a second time, it will be a lot easier if ##y## is canceled out before. This also reduces the risk of making mistakes. Since ##y## must not be zero anyway, it may be canceled.
     
  13. Sep 16, 2016 #12
    There is a mistake in yours 2 step.
    $$\because\;\dfrac{d}{dx}(x)=1\;\&\;\dfrac{d}{dx}(y)^2=2y\cdot\dfrac{dy}{dx}$$
    So, your 2 step will be as-
    $$\dfrac{d}{dx}\left(\dfrac{x}{y^2}\right)=\dfrac{1\cdot y^2-x\cdot2y\dfrac{dy}{dx}}{y^4}$$
    $$=\dfrac{y^2-2xy\left(\dfrac{dy}{dx}\right)}{y^4}$$
    I think this will help you in learning implicit differentiation.
    http://www.actucation.com/calculus-...ative/implicit-derivative-and-its-application
     
  14. Sep 16, 2016 #13
    Yes you're right. Thank you for the correction.
     
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