1. ## logarithmic differentiation

if y=x^cosx, find dy/dx.

help please...Im not sure if I did it right.

2. Originally Posted by edge
if y=x^cosx, find dy/dx.

help please...Im not sure if I did it right.
"Im not sure if I did it right" means you have done something. Show what you did and we can tell you whether it is right or wrong.

3. lny=lnx^cosx
lny=cosxlnx
y'/y=-sinxlnx+cosx(1/x)
y'=y(-sinxlnx+cosx/x)
y'=x^cosx (-sinxlnx+cosx/x)

is that right...and can I simplify it anymore?

4. Originally Posted by edge
lny=lnx^cosx
lny=cosxlnx
y'/y=-sinxlnx+cosx(1/x)
y'=y(-sinxlnx+cosx/x)
y'=x^cosx (-sinxlnx+cosx/x)

is that right...and can I simplify it anymore?
Do you HAVE to use logarithmic differentiation? You can just use chain rule...

$y = x^{\cos{(x)}}$.

Let $u = \cos{(x)}$ so that $y = x^u$.

$\frac{du}{dx} = -\sin{(x)}$

$\frac{dy}{du} = ux^{u - 1} = \cos{(x)}\,x^{\cos{(x)}-1}$

So $\frac{dy}{dx} = -\sin{(x)}\cos{(x)}\,x^{\cos{(x)}-1}$.

5. I'm not sure, does what I did work though?

6. Originally Posted by edge
I'm not sure, does what I did work though?
It looks fine.

Perhaps you'd like to spend some time proving that the two answers are indeed equivalent...

7. Originally Posted by Prove It
Do you HAVE to use logarithmic differentiation? You can just use chain rule...

$y = x^{\cos{(x)}}$.

Let $u = \cos{(x)}$ so that $y = x^u$.

$\frac{du}{dx} = -\sin{(x)}$

$\frac{dy}{du} = ux^{u - 1}$
No!!! This is only true for u a constant!

$= \cos{(x)}\,x^{\cos{(x)}-1}$

So $\frac{dy}{dx} = -\sin{(x)}\cos{(x)}\,x^{\cos{(x)}-1}$.

8. Originally Posted by Prove It
It looks fine.

Perhaps you'd like to spend some time proving that the two answers are indeed equivalent...
He can't- they are not the same. His answer is correct, yours is wrong.