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Need help with a derivative

  1. Jul 14, 2012 #1
    Lets say we have this:
    y=(x^x)^x
    what is y'=?

    I can handle the variable to a variable but am not sure how to approach it when another variable is nested inside the function
     
  2. jcsd
  3. Jul 14, 2012 #2
    You can write the expression as [itex]y=x^{x^{2}}[/itex], take logarithm on both the sides and find the derivative.
     
  4. Jul 14, 2012 #3
    Sorry I was trying to put it in more generic form, lets try it this way:
    y=(3x^2x+1)^x+1

    Great answer by the way!
     
  5. Jul 14, 2012 #4
    You can still do the same, take logarithm on both sides.
     
  6. Jul 14, 2012 #5
    Thats where I am getting stuck, how do I take the ln of both sides with the inner function also being a variable to a variable?
     
  7. Jul 14, 2012 #6
    Oops sorry misread the question.

    When you take logarithm of both the sides, you get an equation like this:
    [tex]\ln y=(x+1)\ln (3x^{2x}+1)[/tex]
    Differentiate it using the product rule, when its turn to differentiate 3x^(2x), calculate its derivative separately and use that.
     
  8. Jul 14, 2012 #7
    Which gets me to here:

    (1/y)(dy/dx)=(x+1)[1/(3x^2+1)]...

    and now we where I am stuck lol, how do I differentiate (3x^2+1) by itself?
     
  9. Jul 14, 2012 #8
    I thought you asked about 3x^(2x).:rolleyes:

    What's the derivative of [itex]y=3x^{2x}[/itex]? :wink:
     
  10. Jul 14, 2012 #9
    Hah ha, yes I did :)
    The derivative of y=3x^2x?
    Don't know
     
  11. Jul 14, 2012 #10
    You know it, don't you? Take logarithm on both sides again.
     
  12. Jul 14, 2012 #11
    So we can just set up a new function for the 3x^2x and set equal to y then differentiate?

    So it would be:

    (1/y)(dy/dx)=(2)+(ln 3x)(2)
     
  13. Jul 14, 2012 #12
    No.
    Let's first calculate the derivative z=3x^(2x) (i use z because we have already used y).
    Differentiating y w.r.t. x, we get:
    [tex]\frac{1}{y}\frac{dy}{dx}=\ln (3x^{2x}+1)+(x+1)\frac{1}{3x^{2x}+1}\frac{d}{dx}(3x^{2x}+1)[/tex]

    Calculate derivative of 3x^(2x) or derivative of z separately.
     
  14. Jul 14, 2012 #13
    Mind if we get rid of that +1 from the original eqation? It was meant to be part of the exponent but I miswrote it

    So we get:
    z=3x^(2x)

    lnz=(2x)ln(3x)
    differntiate to:
    (1/z)(dz/dx)=(2x)(1/(3x)(3)+(2)ln(3x)
     
  15. Jul 14, 2012 #14
    Use the Chain Rule and remember that d/dx(ax) = axln(a) where a is a constant
     
  16. Jul 14, 2012 #15
    Yeah, I was able to get that far but how do we differentiate when there is a variable to a variable power inside a variable to a variable power function?
     
  17. Jul 14, 2012 #16

    Ray Vickson

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    I don't know what your expression means. Is it
    [itex] (3 x^2 x + 1)^x + 1, \; (3 x^{2x}+1)^x + 1,\; (3 x^{2x} + 1)^{x+1},[/itex] or something else? If I read it *using standard rules* it means the first. If you mean the last, use brackets: (3 x^(2x) + 1)^(x+1).

    RGV
     
  18. Jul 14, 2012 #17
    yeah, I knida screwed the pooch on that one, it was supposed to be:

    (3x^(2x+1))^(x+1)
     
  19. Jul 14, 2012 #18
    If it's this, then it should be easy to solve.
    You can write this as:
    [itex]y=3x^{2x^2+3x+1}[/itex]
    Take logarithm on both sides and find the derivative.
     
  20. Jul 14, 2012 #19
    Yup, but I need a solution without simplification
     
  21. Jul 14, 2012 #20

    SammyS

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    Staff Emeritus
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    Gold Member

    Keep applying derivative rules including the chain rule, from the "outside" towards the "inside".

    For instance:

    [itex]\displaystyle \frac{d}{dx}\left(f\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\right)[/itex]
    [itex]\displaystyle =
    f'\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot
    \frac{d}{dx}\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right) [/itex]

    [itex]\displaystyle =
    f'\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot
    g'\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\cdot
    \frac{d}{dx}\left(h\left(s\left(u(x)\right)+v(x) \cdot w(x)\right)\right) [/itex]

    [itex]\displaystyle =
    f'\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot
    g'\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\cdot
    h'\left(s\left(u(x)\right)+v(x) \cdot w(x)\right)\cdot
    \frac{d}{dx}\left(s\left(u(x)\right)+v(x) \cdot w(x)\right) [/itex]

    [itex]\displaystyle =
    f'\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot
    g'\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\cdot
    h'\left(s\left(u(x)\right)+v(x) \cdot w(x)\right)\cdot
    \left(
    s'\left(u(x)\right)\cdot\frac{d}{dx}u(x)+
    \frac{d}{dx}\left(v(x) \cdot w(x)\right)\right) [/itex]


    [itex]\displaystyle =
    f'\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot
    g'\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\cdot
    h'\left(s\left(u(x)\right)+v(x) \cdot w(x)\right)\cdot
    \left(s'\left(u(x)\right)\cdot u'(x)+ \left(v'(x) \cdot w(x)+v(x) \cdot w'(x)\right)\right)[/itex]
     
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