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Differentiating 2^(x^2)

  1. Apr 17, 2012 #1
    I'm trying to differentiate 2^(x^2), but I'm getting a factor of two out and can't figure out why. I approached the question as follows..

    y=2^(x^2) , so y=(2^x)^x
    u=2^x y=u^x

    du/dx = (2^x)ln2
    dy/du = xu^(x-1)
    = x(2^x)^(x-1)
    = x(2)^((x^2)-x)

    So dy/dx =
    [x(2)^((x^2)-x)]*[(2^x)Ln2]

    However, on the mark scheme it says when x=2, the gradient should be 64ln2. Using my derivative, at x=2 the gradient comes out at 32ln2. Can anyone help me find where I've gone wrong? Much appreciated!
     
  2. jcsd
  3. Apr 17, 2012 #2
    [itex]2^{x^2}[/itex] is a composition of functions [itex]f(x) = 2^x[/itex] and [itex]g(x) = x^2[/itex]. You know that [itex]f'(x) = 2^x \ln 2[/itex] and [itex]g'(x) = 2x[/itex]

    There is a formula for derivative of composite functions. [itex](f \circ g)'(x) = f'(g(x))g'(x)[/itex]. This is just another form of the chain rule. After blindly pasting the functions we already have, we get [itex]2x \cdot 2^{x^2} \ln 2[/itex]

    I can't see what error you made as those formulas are not very readable. Try using latex.
     
  4. Apr 17, 2012 #3

    Curious3141

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    That's wrong right there. The rule [itex]\frac{d}{dx}x^n = nx^{n-1}[/itex] ONLY applies when n is a constant. If n is a variable like x or a function of x, the rule simply does not work.

    Your best bet here is to use Chain Rule as hamsterman mentioned.
     
  5. Apr 17, 2012 #4

    hotvette

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    Another approach is to take the (natural) log of both sides of y = 2x2 and differentiate implicitly.
     
  6. Apr 18, 2012 #5
    Thanks guys, got it!
     
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