Differentiate cosecx

**nath_quam** · June 12th 2006, 11:38 PM

Prove that d/dx (cosecx) = -cosecxcotx

**topsquark** · June 13th 2006, 03:53 AM

Originally Posted by nath_quam

Prove that d/dx (cosecx) = -cosecxcotx

Hint: start with this
cscx = 1/sinx

-Dan

**CaptainBlack** · June 13th 2006, 04:13 AM

What would you like us to do with the attachment?

RonL

**Jameson** · June 13th 2006, 05:13 AM

For part (a) integrate dI and then using your initial value solve for the constant of integration.

**ThePerfectHacker** · June 13th 2006, 10:47 AM

Since,
$\csc x=\frac{1}{\sin x}$
Then, when you take its derivative apply quotient rule,
$(\csc x)'=\frac{(1)'\sin x+1(\sin x)'}{\sin^2 x}$
Then,
$(\csc x)'=\frac{0\cdot \sin x-1\cdot \cos x}{\sin^2 x}=\frac{-\cos x}{\sin ^2x}$ = $\frac{-\cos x}{\sin x}\cdot \frac{1}{\sin x}=-\cot x\sin x$
---
For reference,
If $f(x)$ is a diffrenciable function at $x=c$ and $g(x)$ is a diffrenciable function at $x=c$ such as, $g(c)\not = 0$ , then $f(x)/g(x)$ is diffrenciable at $x=c$ and its derivative at $x=c$ is,
$\frac{f'(c)g(c)-f(c)g'(c)}{[g(c)]^2}$

**nath_quam** · June 13th 2006, 01:29 PM

Perfect Hacker shouldn't the quotient rule have a minus in the middle not a plus as you have it Thanks Nath and how would i go about doing the three parts of the attachment

**ThePerfectHacker** · June 13th 2006, 01:53 PM

Originally Posted by nath_quam

not a plus as you have it

A plus!
Where do you see a plus?

(Okay I admit I edited my message, you are right)

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