Carman3
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In the book there's a question which says find the maclaurins series for lncosx using differentiation and I'm finding it difficult to do I searched Google for a way on how to do it but it seems something beyond a level?
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ashmcm
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Put cosx into its exponential form and then use the Maclaurin series for ex maybe?
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ashmcm
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Or just use the generic formula of
Image
where f(x) = ln(cosx), f'(x) = -tanx and f''(x) = -sec2 x
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Carman3
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(Original post by ashmcm)
Put cosx into its exponential form and then use the Maclaurin series for ex maybe?
How would I go about doing that
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davros
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(Original post by Carman3)
In the book there's a question which says find the maclaurins series for lncosx using differentiation and I'm finding it difficult to do I searched Google for a way on how to do it but it seems something beyond a level?
What are you finding difficult? Presumably you know what the definition of a Maclaurin series is, and how to find the coefficients by repeated differentiation...
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the bear
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there is an expansion for ln( 1 + x )

and the expansion of cos begins 1 - x2/2
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Carman3
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(Original post by davros)
What are you finding difficult? Presumably you know what the definition of a Maclaurin series is, and how to find the coefficients by repeated differentiation...
If I let f (x) = lncosx, how do I find f'(x). The question says use differentiation so I have to find f (x)

Perhaps u shouldve been more exact on what I needed help with
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ashmcm
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(Original post by Carman3)
If I let f (x) = lncosx, how do I find f'(x). The question says use differentiation so I have to find f (x)

Perhaps u shouldve been more exact on what I needed help with
Chain rule, differentiating ln(cosx) = 1/cosx * -sinx = -tanx
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Carman3
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Hm. I guess I used to do it differently in core 3 whether that may be wrong. Sorry if it was painfully obvious and thanks all for the help
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