Log Differentiation: Find dy/dx of (cosx)^sinx

[bengalibabu]

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

any help wud be appreciated, I am unsure of how to start this question, thanks in advance

 

[dextercioby]

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]

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

 

[Curious3141]

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

 

[Pyrrhus]

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 I am on the right track. thanks for your help guys.

 

[Pyrrhus]

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]

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"