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Simple maths natural logs question

Problem solved.
(edited 11 years ago)
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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Notnek
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The derivative of ln(f(x)) is the derivative of f(x) divided by f(x):

ddx(ln⁡f(x))=f′(x)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)?
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?
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.
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 :smile:
Original post by notnek
The derivative of ln(f(x)) is the derivative of f(x) divided by f(x):

ddx(ln⁡f(x))=f′(x)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 :smile:

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