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Thread: 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:

    $\displaystyle y=x^\tan(x) $

    $\displaystyle ln y= tan(x) ln x $

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

    Then multiply with the original y to get:

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

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

    Any help would be appreciated.
    Last edited by Bean; Sep 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:

    $\displaystyle y=x^\tan(x) $

    $\displaystyle ln y= tan(x) ln x $

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

    Then multiply with the original y to get:

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

    But the answer I'm seeing has $\displaystyle 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:

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

    Then, use the chain rule to find the derivative:

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

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

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

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

    These are equivalent because $\displaystyle a^{b+c} = a^b \cdot a^c$.
    Last edited by SworD; Sep 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 $\displaystyle 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:

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

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

    $\displaystyle \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:

    $\displaystyle \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: $\displaystyle x^{\tan x-1} $ in there instead of the original y value.

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

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

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

    "Subtract exponents", remember?
    Thanks from MarkFL
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