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Quick derivative question

  1. Jan 10, 2004 #1
    How do you find the derivative of y=x^y ?
  2. jcsd
  3. Jan 10, 2004 #2
    y&=x^y \\
    \ln y&=y\ln x \\
    \frac{y^\prime}{y}&=y^\prime\ln x+\frac{y}{x} \\
    y^\prime\left(\frac{1}{y}-\ln x\right)&=\frac{y}{x} \\
    y^\prime&=\frac{y^2}{x-xy\ln x}
  4. Jan 10, 2004 #3
    Not quite the answer I was looking for, but thanks any how. I already know how the problem is solved.
  5. Jan 10, 2004 #4
    You asked me how to find the derivative. I showed you a way to do it. What were you looking for?
  6. Jan 11, 2004 #5
    I was looking for something more like this: x'=y^(1/y)*(1/y^2-ln(y)/y^2).

    I solved it though, guess just got stuck for a minute.
  7. Jan 11, 2004 #6
    Usually people want the derivative of y wrt x. Plus, you gave an equation for y in terms of x and y.
  8. Jan 11, 2004 #7
    Since y does not equal f(x), y can not be expressed in terms of x. We would have to express x in terms of y.

    Of course, it would be much better if the equation was x=y^x, then it can be expressed as y=x^(1/x).
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