1. ## differentation

differentate :ln(sinx cos x) answer provided2 cot 2x)

dy/du = 1/u du/dx= -sinx cosx

-sin x cos x/( ln [(sinx cosx)] )

-sinx cosx / (1/sinx + 1/cos x )

stucked

2. Best way to start is by using the rule of logs:

$y = \ln \left({\sin x \cos x}\right) = \ln \left({\sin x}\right) + \ln \left({\cos x}\right)$

Diffing that is easy, leads to

$\frac {dy}{dx} = \frac {\cos x}{\sin x} - \frac {\sin x} {\cos x}$

I left it as sin & cos to make the next stage easier, i.e. put it over a common denominator.

Then you have an expression on the top and the bottom for which the double angle formulae for sin 2x and cos 2x can be used, and you're done.

3. did you mean

cos^2 x - sin^2 x/ cos x sin x

so i stucked at here.

btw cos x sin x mean wht? cos^2 x - sin^2 x = cos 2x

4. Originally Posted by qweiop90
did you mean

cos^2 x - sin^2 x/ cos x sin x

so i stucked at here.

btw cos x sin x mean wht? cos^2 x - sin^2 x = cos 2x
So you got the formula for $\cos 2x$ - now what's the formula for $\sin 2x$ then?

5. 2 sinx cos x

6. so where's the problem?

7. 2 sin x cos x is not equivalent to sin x cos x. lets say sin x cos x = sin x

cos 2x / sinx , izzit possible to be 2cot2x?

8. uh?!?!?

Do you know how to multiply and divide by 2?

$\sin 2x = 2 \sin x \cos x = 2 \cos x \sin x$

Theeeerreefoooooooore...

$\cos x \sin x = \frac {\sin 2x}{2}$

9. hehehe ok . i get it. thx