# 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

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dextercioby
Homework Helper
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
Homework Helper
$$a = e^{\ln{a}}$$ where 'ln' is natural log.

Curious3141
Homework Helper
Then use chain rule plus product rule. The other way is to differentiate implicitly after taking logs of both sides.

Pyrrhus
Homework Helper
Apply logarithm on both sides, and use the property of logs $\ln a^b = b \ln a$

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
Homework Helper
I think so, but why this ln 10??

$$\ln y = \sin x \ln (\cos x)$$

$$\frac{dy}{dx} \frac{1}{y} = \sin x \frac{-\sin x}{\cos x} + \cos x \\ln (\cos x)$$

$$\frac{dy}{dx} = (\cos x)^{\sin x}(\sin x \frac{-\sin x}{\cos x} + \cos x \ln (\cos x))$$

isnt the derivative of log(base)x = 1/xln(base) ????

Pyrrhus
Homework Helper
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.

hah - "logarithmate"? I would prefer "logarithmize"