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Simple maths natural logs question

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    Problem solved.
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    If y=ln(f(x)) then dy/dx=f'(x)/f(x)
    In this case f(x) is sinx and f'(x) is cosx therefore dy/dx=cos(x)/sin(x) or cot(x).
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    The derivative of ln(f(x)) is the derivative of f(x) divided by f(x):

    \displaystyle \frac{d}{dx}\left(\ln f(x)\right) = \frac{f'(x)}{f(x)}

    So what is dy/dx where f(x)=sin(x)?
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    chain rule buddy

    differentiate what's inside the outer function so differentiate ln(sinx) to give 1/sinx then multiply by the derivative of the inner function (sinx))

    what's the derivative of sinx?
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    (Original post by kingkongjaffa)
    chain rule buddy

    differentiate what's inside the outer function so differentiate ln(sinx) to give 1/sinx then multiply by the derivative of the inner function (sinx))

    what's the derivative of sinx?
    Oh yeah, thanks. I was trying to use the product rule to differentiate instead.
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    (Original post by Augmented hippo)
    If y=ln(f(x)) then dy/dx=f'(x)/f(x)
    In this case f(x) is sinx and f'(x) is cosx therefore dy/dx=cos(x)/sin(x) or cot(x).
    thank you
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    (Original post by notnek)
    The derivative of ln(f(x)) is the derivative of f(x) divided by f(x):

    \displaystyle \frac{d}{dx}\left(\ln f(x)\right) = \frac{f'(x)}{f(x)}

    So what is dy/dx where f(x)=sin(x)?
    Thanks

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