derivative of cot x

saweetness · #1 $Old$ November 7th, 2009, 04:20 PM

use the quotient rule to show that
d/dx cotx = -csc^2 x
(hint: Write cotx = cosx/sinx)

...

cotx = cosx/sinx

:S

please help THANKS!

Debsta · #2 $Old$ November 7th, 2009, 04:22 PM

Do you know the quotient rule?

e^(i*pi) · #3 $Old$ November 7th, 2009, 04:23 PM

Quote:

Originally Posted by saweetness $View Post$

use the quotient rule to show that
d/dx cotx = -csc^2 x
(hint: Write cotx = cosx/sinx)

...

cotx = cosx/sinx

:S

please help THANKS!

The quotient rule says that:
$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - g'(x)f(x)}{[g(x)]^2}$

$f(x) = cos(x)$
$g(x) = sin(x)$

I-Think · #4 $Old$ November 7th, 2009, 04:24 PM

Quotient rule:
If $f(x)=\frac{p(x)}{q(x)}$
Then $f^|(x)=\frac{q(x)p'(x)-p(x)q'(x)}{[q(x)]^2}$

saweetness · #5 $Old$ November 7th, 2009, 04:31 PM

ok so that would be equal

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

now do i use the proof identities? like

cos^2x - sin^2x = 2cos^2x - 1
and then
sin^2x = 1 - cos^2x

Debsta · #6 $Old$ November 7th, 2009, 04:41 PM

saweetness · #7 $Old$ November 7th, 2009, 04:43 PM

huh?

Debsta · #8 $Old$ November 7th, 2009, 04:50 PM

sorry,
Check the quotient rule again from the start...you seem to have lost a neg sign in front of cos^2 X

Yes and then use the identity sin^2 x + cos^2 x =1

saweetness · #9 $Old$ November 7th, 2009, 05:00 PM

ahhhhhh thanks

so then the top would just equal -1 because of the pythag identity right?

therefore it would be -1/sin^2x or -csc^2x

haha i often forget to pay attention to my signs. thanks alot!

Debsta · #10 $Old$ November 7th, 2009, 05:06 PM

No worries,,,glad I helped.