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Implicit differentiation

  1. Jun 19, 2008 #1
    Im having trouble solving these two questions. I don't know where to start so I cant give an attempt at either of them. Please tell me how to do the full question if you can cause i cant check back until morning then i have to go to an exam.

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  2. jcsd
  3. Jun 19, 2008 #2
    First differentiate the LHS remembering that a dx will pop up then the same for the RHS then divide to find dy/dx.
     
  4. Jun 19, 2008 #3
    Simply differentiate both sides like you are used to, but keep in mind that y is not a constant! The derivative of y is [tex]\frac{dy}{dx}[/tex] so you should leave that in the equation. Then you will find an equation which you can solve for dy/dx.

    I don't know what dirk_mec means by 'a dx will pop up' though. Perhaps take a look at the following example (to not give away the answer I use a different example):

    [tex]y \sin x = x^3 + \cos y[/tex]
    [tex]\frac{d}{dx}(y \sin x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(\cos y)[/tex]
    [tex](\sin x)\frac{dy}{dx} + y \cos x = 3x^2 - (\sin y) \frac{dy}{dx}[/tex]
    [tex](\sin x + \sin y) \frac{dy}{dx} = 3x^2 - y \cos x[/tex]
    [tex]\frac{dy}{dx} = \frac{ 3x^2 - y \cos x}{\sin x + \sin y}[/tex]

    Note the use of the product rule on the LHS in the third line!


    This is called implicit differentiation. Often you can rewrite a function to read 'y = f(x)' (where f is some function). But in this case this is impossible, so you have to use implicit differentiation.
     
    Last edited: Jun 19, 2008
  5. Jun 19, 2008 #4
    Here are my attempts:
    x^2=ln(x+y)
    2x=(1/x)(1/y)dy/dx
    2(x^2)y=dy/dx

    e^(xy)=x+4
    u=xy
    u'=(1)(1)(dy/dx)
    (1)(1)(dy/dx)e=1
    (dy/dx)=1/e

    They're both wrong.
     
  6. Jun 19, 2008 #5

    Mute

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    [tex]\ln(x+y) = \frac{1}{x +y}\left(1 + \frac{dy}{dx}\right)[/tex]

    In general, the derivative of [itex]\ln f(x)[/itex] is [itex]\frac{1}{f(x)}\frac{df}{dx}[/itex], as per the chain rule. In this case, f(x) = x + y(x) (y is a function of x).

    This is just an application of the chain rule again, keeping in mind that y is a function of x. There's no need to introduce u.

    [tex]\frac{d}{dx}e^{xy} = e^{xy}\frac{d}{dx}(xy) = e^{xy}\left(y + x\frac{dy}{dx}\right)[/tex]
     
  7. Jun 19, 2008 #6
    So for ln(x+y)
    its 1/(x+y) multiplied by the derivative of what's inside the parenthesis?
     
  8. Jun 20, 2008 #7
    Yes.

    And for [tex]e^{xy}[/tex] it's [tex]\frac{d}{dx} e^{xy} = e^{xy} \frac{d}{dx}(xy)[/tex]
    (don't forget the product rule!)

    So in words, the derivative of exp(f(x)) is simply exp(f(x)) multiplied by the derivative of f(x).


    So to get you started with the second one, this is the first step:
    [tex]e^{xy} = x + 4[/tex]
    [tex]e^{xy} \frac{d}{dx}(xy) = \frac{d}{dx}(x+4)[/tex]
    [tex]e^{xy} \left( y + x \frac{dy}{dx} \right) = 1[/tex]
    ...
     
    Last edited: Jun 20, 2008
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