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Another implicit differentiation

  1. Mar 6, 2007 #1
    e^x^y = x +y

    ok i know i am suppost to use the chain rule and the product rule

    so x+y is 1 +1 if u find the derivatives, but e^x^2 is confusing me, what is u and what is n

    i think u= e^x^2 and n= y is that possible for n to equal y, this problem is confusing
     
  2. jcsd
  3. Mar 7, 2007 #2

    dextercioby

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    So you need to find [itex] \frac{dy}{dx} [/itex]. Assume

    [tex] F(x,y)=e^x^y-x-y [/tex]

    What are
    [tex] \frac{\partial F}{\partial x} \ , \frac{\partial F}{\partial y} [/tex]

    equal to ?
     
  4. Mar 7, 2007 #3

    HallsofIvy

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    Your notation is ambiguous. Is e^x^2 supposed to be (e^x)^2 or e^(x^2). I would guess the latter since (e^x)^2 can be written more simply as e^(2x) but you should make it clear.

    Oh, and what in the world do you mean by "so x+ y is 1+ 1"? If you mean "the derivative of x+ y is 1+ 1", that is wrong. What variable are you differentiating with respect to? Since you mention the "chain rule" I would guess that x and y are functions of some third variable- again you should tell us that and not make us guess. (Okay, I now notice that the title of this was "another implicit differentiation! But we still don't know whether we are to assume y is a function of x or vice-versa.)

    I surely can't answer your question "what is n" because there is no "n" in your formula! Again, I can guess that you mean the "n" in the formula (xn)'= n xn-1 but I can't be sure.

    PLEASE restate the problem exactly as it given! (And don't just tell us it is "e^x^y= x+y". That's not even a problem. It may well be one of a list of problems in which the "instructions" [what you are to assume and what you are to do] are at the top of the list.)

    It might help you to think about how you would differentiate xx or, more generally, xf(x), since y is a function of x. (Do you remember "logarithmic differentiation"?)
     
    Last edited: Mar 7, 2007
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