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Differentiation With 'x' Both in Base and in Exponent

  1. Jul 12, 2009 #1
    I have just begun studying differentiaition and I was getting confused with how to differentiate a function of x in which x is in the base and is in the exponent as well (for example, x^2x) with respect to a=hte cahnge in 'x'.
    I remember my teacher telling me something about applying log, but, I got kinda drowsy... I wasn't able to learn it properly... now I need help.... hehehe...
  2. jcsd
  3. Jul 12, 2009 #2
    The easiest way that I know how to find this derivative is to re-write:
    [tex]x = e^{\ln(x)}[/tex]

    So we have:
    [tex]f(x) = x^{2x} = e^{2x\ln x}[/tex]

    [tex]f'(x) = \dfrac{d}{dx}e^{2x\ln x} = e^{2x\ln x}\dfrac{d}{dx}(2x\ln x) = x^{2x}(\frac{2x}{x} + 2\ln x) = 2x^{2x}(1 + \ln x)[/tex]
    Last edited: Jul 12, 2009
  4. Jul 12, 2009 #3


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    pbandjay use the exponential form. What your teacher was talking about was this:
    to differentiate x2x, write y= x2x and take the natural logarithm of both sides: ln(y)= ln(x2x)= 2x ln(x). Now differentiate both sides. On the left, the derivative of ln(y), with respect to x, is (1/y) y'. On the right, use the product rule: (2x ln(x))'= (2) ln(x)+ (2x)(1/x)= 2ln(x)+ 2. So (1/y)y'= 2ln(x)+ 2 and y'= (2ln(x)+ 2)y= (2ln(x)+ 2)x2x.

    A rather amusing point: There are two obvious mistakes one could make in differentiating f(x)g(x):

    1) Treat g(x) as if it were a constant. Then you get (f(x)g(x))'= g(x)(f(x))g(x)-1f'(x)- which is, of course, wrong.

    2) Treat f(x) as if it were a constant. Then you get (f(x)g(x))'= f(x)g(x) ln(f(x))g'(x)- which is also wrong.

    To do it correctly, we can let y= f(x)g(x) so that ln(y)= ln(f(x)g(x))= g(x)ln(f(x) and differentiate both sides, using the product rule on the right.

    (1/y) dy/dx= g'(x)ln(f(x)+ (1/f(x))g(x)f'(x) and, since y= f(x)g(x),
    dy/dx= (f(x)g(x))'= f(x)g(x)ln(f(x)g'(x)+ f(x)g(x)-1f'(x),

    the sum of the two mistaken derivatives!
  5. Jul 12, 2009 #4
    part of the general method called "the chain rule for partial derivatives" which you (the OP) will meet later.

    Another one:
    Two mistaken ways to differentiate a product [tex]f(x) g(x)[/tex] ... in one you think that [tex]f(x)[/tex] is constant, in the other you think that [tex]g(x)[/tex] is constant. BUT the product rule, the correct derivative, is just the sum of these two mistaken derivatives.
  6. Jul 12, 2009 #5
    One problem I have always liked a lot that is very similar to this one is

    To differentiate we can re-write: y = xy and apply the logarithm.

    Re-write: ln(y) = y ln(x)
    Differentiate: y'/y = y/x + y' ln(x)
    Re-arrange: y'/y - y' ln(x) = y/x ==> y'(1/y - lnx) = y/x ==> y'(1 - y lnx) = y2/x
    Isolate y': y' = y2/[x(1 - y lnx)]

    We have
    [tex]y'= \frac{(x^{x^{x^{.^{.^{.}}}}})^2}{x(1-(x^{x^{x^{.^{.^{.}}}}})\ln x)}[/tex]
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