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Math Help - Derivative of x^tan(x)

  1. #1
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    Derivative of x^tan(x)

    I thought I knew how to do this but I keep coming up with the wrong answer. Here is what I've done:

     y=x^\tan(x)

     ln y= tan(x) ln x

     \frac{1}{y}*\frac{dy}{dx}= \tan(x)*\frac{1}{x} +ln x*sec^2 x

    Then multiply with the original y to get:

     \frac{dy}{dx}= x^{\tan(x)} * (\frac{tan(x)}{x}+ ln x*sec^2 x)

    But the answer I'm seeing has  x^{\tan(x)-1} in there instead of the original y value.

    Any help would be appreciated.
    Last edited by Bean; September 12th 2012 at 05:35 PM.
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  2. #2
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    Re: Derivative of x^tan(x)

    Quote Originally Posted by Bean View Post
    I thought I knew how to do this but I keep coming up with the wrong answer. Here is what I've done:

     y=x^\tan(x)

     ln y= tan(x) ln x

     \frac{1}{y}*\frac{dy}{dx}= \tan(x)*\frac{1}{x} +ln x*sec^2 x

    Then multiply with the original y to get:

     \frac{dy}{dx}= x^{\tan(x)} * (\frac{tan(x)}{x}+ ln x*sec^2 x)

    But the answer I'm seeing has  x^{\tan(x)-1} in there instead of the original y value.

    Any help would be appreciated.
    It looks to me that if we take a 1/x out of both terms then you would have the correct exponential. But since there are two terms in the parenthesis you'd have to take the 1/x out of both terms.

    -Dan
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    Re: Derivative of x^tan(x)

    I would suggest that instead of taking the log, you change the function to the following form:

     y = e^{tan(x) \cdot ln(x)}, which is equivalent because a^{bc} = (a^b)^c.

    Then, use the chain rule to find the derivative:

     y' = e^{tan(x) \cdot ln(x)} \cdot \left (\frac{\tan(x)}{x} + \sec^2(x)\cdot \ln(x)\right )

    y' = x^{\tan(x)} \cdot \left (\frac{\tan(x)}{x} + \sec^2(x)\cdot \ln(x)\right )

    Edit: actually, your answer is correct. Perhaps the reason for the discrepancy is that they distributed and changed

    x^{\tan(x)} \cdot \frac{\tan(x)}{x}
    = x^{\tan(x)-1} \cdot \tan(x)

    These are equivalent because a^{b+c} = a^b \cdot a^c.
    Last edited by SworD; September 12th 2012 at 06:05 PM.
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    Re: Derivative of x^tan(x)

    Oh right. I see now that the second term was changed to  x sec^{2} x ln x in the solution. Thanks, guys.
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    Re: Derivative of x^tan(x)

    Hello, Bean!

    I thought I knew how to do this but I keep coming up with the wrong answer.

    Here is what I've done:

     y\:=\:x^\tan(x)

     \ln y\:=\: \tan(x) \ln x

     \frac{1}{y}\cdot\frac{dy}{dx}\:=\: \tan x\cdot \frac{1}{x} + \ln x\cdot\sec^2 x

    Then multiply with the original y to get:

     \frac{dy}{dx}\:=\: x^{\tan x}\cdot \left(\tfrac{\tan x}{x}+ \ln x\!\cdot\!\sec^2 x\right) . This is correct!


    But the answer I'm seeing has:  x^{\tan x-1} in there instead of the original y value.

    They did something sneaky . . . (legal but sneaky!)

    We have: . \frac{dy}{dx} \;=\;\underbrace{x^{\tan x}\!\cdot\!\frac{\tan x}{x}} \;+\; x^{\tan x}\!\cdot\!\ln x\!\cdot\!\sec^2x

    . . Note that: . x^{\tan x}\!\cdot\!\frac{\tan x}{x} \;=\;\frac{x^{\tan x}}{x}\!\cdot\!\tan x \;=\; x^{\tan x -1}\!\cdot\!\tan x

    "Subtract exponents", remember?
    Thanks from MarkFL
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