# Log differentiation

bengalibabu
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

## Answers and Replies

Homework Helper
Logarithmate and differentiate using the chain rule.

Daniel.

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

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

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

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

bengalibabu
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.

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

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

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

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

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

Homework Helper
Yes, you should have used natural logarithm to avoid that, but it's ok.

bengalibabu
how could u have used natural logs to avoid that?

bengalibabu
o nevermind stupid question lol. i should have converted log to ln then found the deriative.

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