Results 1 to 5 of 5

Math Help - Derivative of x^x

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    60

    Derivative of x^x

    I read the theory over and over again, but i don't get it.
    You want the derivative of x^x. The theory says

    "If , then . The chain rule is then used, to differentiate ln(y) with respect to x (y'/y). The product rule is then used on the right. Then solve for y'"

    So you'd get
    ln (y)'--> (1/y) =ln(x)+1<-- xln(x)'
    Thus y= 1/ln(x)+1, which is wrong. The derivative of x^x should be:x^x(ln(x)+1)
    1) Where is the mistake?
    2)as the graph of y=f(x) is the same as the graph of ln(y)=ln(f(x)), why can you not simply derive the "right side", why do you have to derive both sides? In "normal" deriviation, you also don't derive the y!


    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,686
    Thanks
    617
    Hello, Schdero!

    I can't connect to that link.


    Differentiate: . y \:=\:x^x

    Take logs: . \ln y \:=\:\ln(x^x) \quad\Rightarrow\quad \ln y \:=\:x\ln x

    Differentiate implicitly: . \frac{1}{y}\!\cdot\!\frac{dy}{dx} \:=\:x\!\cdot\!\frac{1}{x} + \ln x

    We have: . \frac{1}{y}\!\cdot\!\frac{dy}{dx} \;=\;1 + \ln x \quad\Rightarrow\quad \frac{dy}{dx} \;=\;y\left(1 + \ln x\right)

    Therefore: . \frac{dy}{dx}\;=\;x^x\left(1 + \ln x\right)

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2010
    Posts
    60
    <br />
\frac{1}{y}\!\cdot\!\frac{dy}{dx}<br />
    why is it necessary, do derive ln(y) at all? \frac{dy}{dx}<br />
means the right side derived, but why do you also have to derive the ln(y)?

    PS: sry about the link
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by Schdero View Post
    <br />
\frac{1}{y}\!\cdot\!\frac{dy}{dx}<br />
    why is it necessary, do derive ln(y) at all? \frac{dy}{dx}<br />
means the right side derived, but why do you also have to derive the ln(y)?

    PS: sry about the link
    I'm not exactly sure what your problem is. There seem to be two issues involved. One is how to determine the derivative y'(x) of y(x) :=x^x with respect to x. You can do this without reference to the "rule of logarithmic differentiation", using chain rule and product rule only, like this:

    y'(x)=\left(x^x\right)'=\left(\mathrm{e}^{x\cdot \ln x}\right)'=\mathrm{e}^{x\cdot\ln x}\cdot (1\cdot \ln x+x\cdot \tfrac{1}{x})=x^x\cdot (\ln x+1)

    The other issue is the "rule of logarithmic differentiation" which says that y'(x)=y(x) \cdot\left(\ln y(x)\right)'.
    This follows from differentiation of \ln y(x) with respect to x by the chain rule: \left(\ln y(x)\right)' =\frac{1}{y(x)}\cdot y'(x). Solving for y'(x) gives the above rule.

    And, yes, differentiation of y(x)=x^x can be done by referring to this rule of logarithmic differentation (if you happen to know it):
    y'(x)=y(x)\cdot\left(\ln y(x)\right)' =x^x\cdot \left(\ln x^x\right)' = x^x\cdot \left(x\cdot \ln x\right)'=x^x\cdot (\ln x+1)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2010
    Posts
    60
    I didn't see, respectively understand, the thing with the chain rule, but it's clear now. thanks for both your patience and explenations ;-)
    Schdero
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. contuous weak derivative $\Rightarrow$ classic derivative ?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 22nd 2011, 02:37 AM
  2. Replies: 0
    Last Post: January 24th 2011, 11:40 AM
  3. [SOLVED] Definition of Derivative/Alt. form of the derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 23rd 2010, 06:33 AM
  4. Derivative of arctan in a partial derivative
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 12th 2010, 01:52 PM
  5. Replies: 2
    Last Post: November 6th 2009, 02:51 PM

Search Tags


/mathhelpforum @mathhelpforum