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Homework Help: Exponential Derivative

  1. Nov 22, 2006 #1
    How do you find the derivative of:

    y=x(3x)^x^2
    ?

    I though it was yprime=(1)3x^x^2+x(3x^x^2)ln(1+3x)(2x)
     
  2. jcsd
  3. Nov 22, 2006 #2
    I got

    1*3x^x^2+x(3x^x^2)*(2x ln(3x) + x).

    I'm guessing you just did some bad algebra. Maybe instead of ln(1+3x) you meant (ln(3x) + 1)?
     
  4. Nov 22, 2006 #3

    arildno

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    You could do it in various ways.
    One method (ultimately justified by the multi-variable chain rule) is to regard one x as varying, the others constant, and then add the results:
    [tex]\frac{dy}{dx}=(3x)^{x^{2}}+3x^{3}(3x)^{x^{2}-1}+2x^{2}(3x)^{x^{2}}\ln(3x)[/tex]

    To do this properly, we may define:
    [tex]H(u,v,w)=uv^{w}, U(x)=x, V(x)=3x, W(x)=x^{2}[/tex]
    Then, we may define the function:
    [tex]h(x)=H(U(x),V(x),W(x))[/tex]
    and we have, by the multi-variable chain rule:
    [tex]\frac{dh}{dx}=(\frac{\partial{H}}{\partial{u}}\frac{dU}{dx}+\frac{\partial{H}}{\partial{v}}\frac{dV}{dx}+\frac{\partial{H}}{\partial{w}}\frac{dW}{dx})\mid_{(u,v,w)=(U(x),V(x),W(x)}[/tex]
     
    Last edited: Nov 22, 2006
  5. Nov 22, 2006 #4
    i thought the law of y=x^x was:

    x^x(1+lnx)
     
  6. Nov 22, 2006 #5
    i was taught taht you should find the log of both sides and differentiate from there

    so then you question
    [tex] y = x(3x)^{x^2} [/tex]
    will turn into

    [tex] \ln y = \ln (x(3x)^{x^2}) [/tex]

    and differentiate both sides from here
     
  7. Nov 23, 2006 #6

    dextercioby

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    There's a name for what you did: logarithmic differentiation. It basically allows one to get the derivative of any (no matter how complicated) elementary function.

    Daniel.
     
  8. Nov 23, 2006 #7

    HallsofIvy

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    Yes, that's perfectly correct.

    Also correct. In particular, if y= xx then ln y= x ln x so
    (1/y)y'= ln x+ x(1/x)= ln(x)+ 1. y'= y(ln x+ 1)= xx(ln x+ 1) as thomasrules said. You can use (xx)'= xx(ln x+ 1) together with the product and chain rules to do this more complicated version:
    [tex]y= x(3x)^{x^2}[/tex]
    then
    [tex]y'= (3x)^{x^2}+ x((3x)^{x^2})(ln(3x)+ 1)(2x)[/tex]
    or you can say
    [tex]\ln y = \ln (x(3x)^{x^2})= ln x+ x^2 ln 3+ x^2 ln(x)[/tex]
    and differentiate that.
     
  9. Nov 23, 2006 #8

    arildno

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    Sure enough, and you can do find this out in various ways:
    1. One variable chain rule
    [tex]y=x^{x}=e^{x\ln(x)}=e^{u(x)},\frac{dy}{dx}=\frac{d}{du}e^{u}*\frac{du}{dx}=e^{u(x)}*(\ln(x)+x*\frac{1}{x})=x^{x}(1+\ln(x))[/tex]
    2. Logarithmic differentiation:
    [tex]Y(x)=\ln(y(x))=x\ln(x),\frac{dY}{dx}=\ln(x)+1=\frac{dY}{dy}\frac{dy}{dx}\to\frac{1}{y(x)}\frac{dy}{dx}=1+\ln(x)\to\frac{dy}{dx}=y(x)(1+\ln(x))=x^{x}(1+\ln(x))[/tex]
    3. Multivariable chain rule (regard one x as constant, the others varying, then add the results)):
    [tex]\frac{dy}{dx}=x*x^{x-1}+x^{x}\ln(x)=x^{x}(1+\ln(x))[/tex]
    As noted, there are various valid methods to do this correctly. Logarithmic differentiation is ONE technique, among several.
     
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