# Homework Help: Log differentiation

1. Jul 14, 2005

### 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

2. Jul 14, 2005

### dextercioby

Logarithmate and differentiate using the chain rule.

Daniel.

3. Jul 14, 2005

### bengalibabu

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

4. Jul 14, 2005

### Curious3141

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

5. Jul 14, 2005

### Curious3141

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

6. Jul 14, 2005

### Pyrrhus

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

7. Jul 14, 2005

### 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.

8. Jul 14, 2005

### Pyrrhus

I think so, but why this ln 10??

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

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

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

9. Jul 14, 2005

### bengalibabu

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

10. Jul 14, 2005

### Pyrrhus

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

11. Jul 14, 2005

### bengalibabu

how could u have used natural logs to avoid that?

12. Jul 14, 2005

### bengalibabu

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

13. Jul 15, 2005

### Jelfish

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