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When to assign a value to a multiple chain derivative?

  1. Oct 30, 2011 #1
    I was doing my homework and I ran into a problem of a chain rule within a chain rule. When do I know what to assign a value? For example:

    [tex]y=e^{-x^2}[/tex]

    When I assign [tex]u=e^{-x}[/tex] and [tex]y=u^2[/tex] I get a wrong value. According to cramster I was supposed to assign y = e^u and u=-x^2. But when am I supposed to know what value to assign? For clarification let me write out the steps my problem.

    [tex]y=e^{-x^2}[/tex]
    [tex]u=e^{-x} -----> \frac{du}{dx}=-e^{-x}[/tex]
    [tex]y=u^2 ---------> \frac{dy}{du}=2u[/tex]
    This shows out that the chain is consistent.
    [tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}
    [tex]=2u(-e^{-x})[/tex]
    [tex]=2(e^{-x})(-e^{-x})[/tex]
    [tex]=-2e^{-2x}[/tex]

    But cramster has a different answer. Cramster assigned the values as y=e^u and u =-x^2 got an answer of:
    [tex]=-2x^{-x^2}[/tex]
     
    Last edited: Oct 30, 2011
  2. jcsd
  3. Oct 30, 2011 #2

    gb7nash

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    This is your problem. [itex]u^{2}[/itex] is not equal to [itex]e^{-x^2}[/itex]. In fact, it's equal to [itex]e^{-x}e^{-x} = e^{-2x}[/itex]
     
  4. Oct 30, 2011 #3
    No I assign [tex]y=u^2[/tex]. So then [tex]f'(y)=2u[/tex]

    Edit: I think there is a bit of a confusion. This is the answer I got-->[tex]e^{-x} = e^{-2x}[/tex]. I think you confused it with my last line; the last line is the answer posted up on cramster. I edited my original post and clarified it.
     
    Last edited: Oct 30, 2011
  5. Oct 30, 2011 #4

    gb7nash

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    Is this the same y as:

    ?
     
  6. Oct 30, 2011 #5
    Ohhhhh.. so in this case u^2 implies the function [tex](e^{-x})^2[/tex] Where it is supposed to be only [tex]-x^2[/tex]

    That is what your saying right? Because I just noticed.
     
  7. Oct 30, 2011 #6

    gb7nash

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    Correct
     
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