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Math Help - Derivatives of exponential functions

  1. #1
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    Derivatives of exponential functions

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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by ozgunatalay View Post
    This will require the chain rule and using logarithmic differentiation

    y = \frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4}{9}}}{  (x^3+6x)^{\frac{4}{11}}}

    ln(y) = ln \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4}  {9}}}{(x^3+6x)^{\frac{4}{11}}}\right)

    Since ln \left(\frac{ab}{c}\right) = ln(a)+ln(b)-ln(c) we get


    ln(y) = ln \left[(x^2+2)^{\frac{3}{2}}\right] + ln \left[(x^2+9)^{\frac{4}{9}}\right] - ln \left[{(x^3+6x)^{\frac{4}{11}}}\right]

    Use the chain rule for each of them:

    edit: oops no it isn't *edits*
    edit2: that's better


    Spoiler:


    \frac{1}{y} \, y' = \frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x}{11(x^3+6x)^{\frac{4}{11}}}

    y' = y\, \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x}{11(x^3+6x)^{\frac{4}{11}}}\right)

    y' = \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4}  {9}}}{(x^3+6x)^{\frac{4}{11}}}\right) \cdot \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x}{11(x^3+6x)^{\frac{4}{11}}}\right)
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  3. #3
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    Thank you for your solution.
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  4. #4
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    e^(i*pi)'s Avatar
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    I suppose you could always use the quotient rule, the product rule and the chain rule but IMO that would lead to an even bigger mess
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  5. #5
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    I am learning about this new. Sometimes when I'm having difficulty solving questions. Thanks for your suggestions.
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  6. #6
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    You might also want to note that there are NO exponential functions here. "Exponential functions" are functions that have the variable, x, in the exponent.
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  7. #7
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    I looked again and I ask questions, I reached a different result. If I've made a mistake, correct my mistakes.

    Last edited by ozgunatalay; November 29th 2009 at 06:18 AM.
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  8. #8
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    Is there anyone who can check if I did it right?
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  9. #9
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    I can see no difference between your answer and e^(i*pi)'s, so I would conclude that this is correct.

    EDIT: Ahh I see it now. e^(i*pi) did make a slight mistake in the last fraction:

    <br />
y' = \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4}  {9}}}{(x^3+6x)^{\frac{4}{11}}}\right) \cdot \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x}{11(x^3+6x)^{\frac{4}{11}}}\right)<br />

    is supposed to be:


    <br />
y' = \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4}  {9}}}{(x^3+6x)^{\frac{4}{11}}}\right) \cdot \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x^2+24}{11(x^3+6x)^{\frac{4}{11}}}\right)<br />
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  10. #10
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    Where am I doing wrong?
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  11. #11
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    Nowhere. Your answer is the same as the corrected one.
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  12. #12
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  13. #13
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    Sorry , I didn't check out e^(i*pi)'s answer good enough.

    But your answer is correct.
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  14. #14
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    Thank you for your interest in question
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