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Math Help - more help needed

  1. #1
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    more help needed

    find dy/dx of

    y= ln(x^2 + 9)^4

    i know you have to use the chain rule but i really need some help. i've got an exam tomorrow and i don't konw how to do this
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  2. #2
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    Quote Originally Posted by boomersooner1331 View Post
    find dy/dx of

    y= ln(x^2 + 9)^4

    i know you have to use the chain rule but i really need some help. i've got an exam tomorrow and i don't konw how to do this
    Remember the properties of logarithms:

    y=ln(x^2+9)^4=4ln(x^2+9)

    I always think of the inner function as a variable u so that the application of the chain rule is easier. So think of the function as y=4ln(u), where u is a function of x.

    \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

    \frac{dy}{dx}=4\frac{d}{du}ln(u)\frac{du}{dx}

    =\frac{4}{u}(2x)=\frac{8x}{u}=\frac{8x}{x^2+9}
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  3. #3
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    Wink you couldnot do this!?

    y=ln(x^2+9)^4
    =>y=4ln(x^2+9)
    =>dy/dx=4.1/(x^2+9).2x
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  4. #4
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    so because it's a logarithm you don' multiply it by 4, and have 3 as an exponent?
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  5. #5
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    Quote Originally Posted by boomersooner1331 View Post
    so because it's a logarithm you don' multiply it by 4, and have 3 as an exponent?
    One of the properties of logarithms is that ln(x)^a=aln(x). This is just an expression usually introduced in pre-calculus courses. I think that answers your question .
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