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Log differentiation

Using log differentiation, find dy/dx, in terms of x for the following:
y = (cosx)^sinx

any help wud be appreciated, im unsure of how to start this question, thx in advance
 

dextercioby

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Logarithmate and differentiate using the chain rule.

Daniel.
 
all i need help is in how to logarithmate it, if i understood that then i should be able to differentiate it easily
 

Curious3141

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[tex]a = e^{\ln{a}}[/tex] where 'ln' is natural log.
 

Curious3141

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Then use chain rule plus product rule. The other way is to differentiate implicitly after taking logs of both sides.
 

Pyrrhus

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Apply logarithm on both sides, and use the property of logs [itex] \ln a^b = b \ln a [/itex]
 
ok. i think i got it

y = (cosx)^sinx
logy = log(cosx^sinx)
logy=sinx(logcosx)
(dy/dx)(1/(yln10)) = sinx(1/cosx(ln10) + log(cosx^cosx)
(dy/dx)=[(tanx/ln10) + (log(cosx^cosx))](yln10)

jus let me know if im on the right track. thanks for your help guys.
 

Pyrrhus

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I think so, but why this ln 10??

[tex] \ln y = \sin x \ln (\cos x) [/tex]

[tex] \frac{dy}{dx} \frac{1}{y} = \sin x \frac{-\sin x}{\cos x} + \cos x \\ln (\cos x) [/tex]

[tex] \frac{dy}{dx} = (\cos x)^{\sin x}(\sin x \frac{-\sin x}{\cos x} + \cos x \ln (\cos x)) [/tex]
 
isnt the derivative of log(base)x = 1/xln(base) ????
 

Pyrrhus

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Yes, you should have used natural logarithm to avoid that, but it's ok.
 
how could u have used natural logs to avoid that?
 
o nevermind stupid question lol. i should have converted log to ln then found the deriative.
 
138
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hah - "logarithmate"? I would prefer "logarithmize"
 

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