Prove that d/dx (cosecx) = -cosecxcotx

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- Jun 12th 2006, 11:38 PMnath_quamDifferentiate cosecx
Prove that d/dx (cosecx) = -cosecxcotx

- Jun 13th 2006, 03:53 AMtopsquarkQuote:

Originally Posted by**nath_quam**

cscx = 1/sinx

-Dan - Jun 13th 2006, 04:13 AMCaptainBlack
What would you like us to do with the attachment?

RonL - Jun 13th 2006, 05:13 AMJameson
For part (a) integrate dI and then using your initial value solve for the constant of integration.

- Jun 13th 2006, 10:47 AMThePerfectHacker
Since,

$\displaystyle \csc x=\frac{1}{\sin x}$

Then, when you take its derivative apply quotient rule,

$\displaystyle (\csc x)'=\frac{(1)'\sin x+1(\sin x)'}{\sin^2 x}$

Then,

$\displaystyle (\csc x)'=\frac{0\cdot \sin x-1\cdot \cos x}{\sin^2 x}=\frac{-\cos x}{\sin ^2x}$=$\displaystyle \frac{-\cos x}{\sin x}\cdot \frac{1}{\sin x}=-\cot x\sin x$

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For reference,

If $\displaystyle f(x)$ is a diffrenciable function at $\displaystyle x=c$ and $\displaystyle g(x)$ is a diffrenciable function at $\displaystyle x=c$ such as, $\displaystyle g(c)\not = 0$, then $\displaystyle f(x)/g(x)$ is diffrenciable at $\displaystyle x=c$ and its derivative at $\displaystyle x=c$is,

$\displaystyle \frac{f'(c)g(c)-f(c)g'(c)}{[g(c)]^2}$ - Jun 13th 2006, 01:29 PMnath_quamQuotient Rule
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

- Jun 13th 2006, 01:53 PMThePerfectHackerQuote:

Originally Posted by**nath_quam**

Where do you see a plus?

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