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Math Help - log differentiation

  1. #1
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    log differentiation

    Differentiate with respect to 'x'.  log_{4}(x^{2})

    I know that  \frac{d}{dx}(log_a{x}) = \frac{1}{lna} \times \frac{1}{x}

    so, based on this formula, I got the answer:  \frac{1}{ln4} \times \frac{1}{x^{2}}

    But this is not the correct answer, can someone please show me how to correctly differentiate this function?

    Thank you!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Tweety View Post
    Differentiate with respect to 'x'.  log_{4}(x^{2})

    I know that  \frac{d}{dx}(log_a{x}) = \frac{1}{lna} \times \frac{1}{x}

    so, based on this formula, I got the answer:  \frac{1}{ln4} \times \frac{1}{x^{2}}

    But this is not the correct answer, can someone please show me how to correctly differentiate this function?

    Thank you!
    Hint: \log_4(x^2)=2\log_4x.

    Can you continue?

    EDIT: If you continue doing it your way, you have to apply chain rule to the x^2 term.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Tweety View Post
    Differentiate with respect to 'x'.  log_{4}(x^{2})

    I know that  \frac{d}{dx}(log_a{x}) = \frac{1}{lna} \times \frac{1}{x}

    so, based on this formula, I got the answer:  \frac{1}{ln4} \times \frac{1}{x^{2}}

    But this is not the correct answer, can someone please show me how to correctly differentiate this function?

    Thank you!
    You need to apply the chain rule:  \frac{d}{dx}\left(f(g(x))\right) = f'(g(x))\cdot g'(x)

    In this case  f(x) = \log_4(x) and  g(x)=x^2 .

    So  \left(\log_4(x^2)\right)' = \frac{1}{\ln(4)}\frac{1}{x^2}\cdot 2x = \frac{2}{x\cdot \ln(4)}
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    Hint: \log_4(x^2)=2\log_4x.

    Can you continue?

    EDIT: If you continue doing it your way, you have to apply chain rule to the x^2 term.
    I think so....

     2 \times \frac{1}{ln4} \times \frac{1}{x}

     \frac{2}{ln4} \times \frac{1}{x}

     \frac{1}{ln2} \times \frac{1}{x}


    Are you 'allowed' to factor out the '2' like this? Even though there is ln 'attached' to 4?

    Is this method correct?
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Tweety View Post
    I think so....

     2 \times \frac{1}{ln4} \times \frac{1}{x}

     \frac{2}{ln4} \times \frac{1}{x}

     \frac{1}{ln2} \times \frac{1}{x}

    Are you 'allowed' to factor out the '2' like this? Even though there is ln 'attached' to 4?
    You can't do that. :/

    The answer should remain as \frac{2}{x\ln 4}.
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Chris L T521 View Post
    You can't do that. :/

    The answer should remain as \frac{2}{x\ln 4}.
    Actually I think you can...

     \ln(4) = \ln(2^2) = 2\ln(2)
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  7. #7
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    But is my method correct? As I simply used the formula in my book, I did not use the chain rule.
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Actually I think you can...

     \ln(4) = \ln(2^2) = 2\ln(2)
    Oh wow...why didn't I see that? XD

    But yes...that's right... XD
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  9. #9
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Tweety View Post
    But is my method correct? As I simply used the formula in my book, I did not use the chain rule.
    If you used  \log_4(x^2)=2\log_4(x) , then the chain rule is not needed.
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  10. #10
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by chiph588@ View Post
    If you used  \log_4(x^2)=2\log_4(x) , then the chain rule is not needed.
    As a matter of learning, i think it's important to identify that the chain rule is still used! But it's the fact that the derivative of x is equal to 1, which allows us to neglect directly thinking about the chain rule.

    This is an obvious result but I think for those having trouble, it might be important!
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