# 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

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#### dextercioby

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

#### 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 $\ln a^b = b \ln a$

#### 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.

#### Pyrrhus

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

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

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

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

#### bengalibabu

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.

#### 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"

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