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Thread: Derivative of a function

  1. November 15th 2006, 05:18 PM #1
    tttcomrader
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    Derivative of a function

    How do you take the derivative of f(x) = arctan(2^x)?
    Last edited by tttcomrader; November 16th 2006 at 04:39 PM.
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  2. November 15th 2006, 05:30 PM #2
    Jacobpm64
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    well, all you have to know is this:

    \frac{dy}{dx}(arctan(x)) = \frac{1}{1 + x^2}

    Then, you have to apply the chain rule to find the derivative of the inside function x^2.

    So, here is your entire problem.

    f(x) = arctan(x^2)
    f'(x) = \frac{1}{1+(x^2)^2}*2x
    f'(x) = \frac{2x}{1+x^4}
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  3. November 16th 2006, 04:38 PM #3
    tttcomrader
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    Sorry, I put down the wrong problem.

    The correction question is:

    f(x) = arctan(2^x)
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  4. November 16th 2006, 09:13 PM #4
    Jameson
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    Same principle. Let u=2^x.

    Thus \frac{d}{dx} \arctan(u) = \frac{1}{1+u^2} \times \frac{du}{dx}

    Do you know how to find the derivative of u?
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  5. November 17th 2006, 03:00 PM #5
    tttcomrader
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    So do you take the natural log of both sides now?
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  6. November 17th 2006, 03:16 PM #6
    Plato
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    Quote Originally Posted by tttcomrader View Post
    So do you take the natural log of both sides now?
    No. That is not necessary.
    For any a>0 the derivative of a^x is a^x \ln (a).
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  7. November 18th 2006, 11:10 AM #7
    Jameson
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    Yes, but to derive that rule you use natural logs.

    u=2^x

    \ln(u)=x\ln(2)

    \frac{1}{u} \times \frac{du}{dx}=\ln(2)

    \frac{du}{dx}=u\ln(2)
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