1. ## Differentiate cosecx

Prove that d/dx (cosecx) = -cosecxcotx

2. Originally Posted by nath_quam
Prove that d/dx (cosecx) = -cosecxcotx
cscx = 1/sinx

-Dan

3. What would you like us to do with the attachment?

RonL

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

5. 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,
$(\csc x)'=\frac{0\cdot \sin x-1\cdot \cos x}{\sin^2 x}=\frac{-\cos x}{\sin ^2x}$= $\frac{-\cos x}{\sin x}\cdot \frac{1}{\sin x}=-\cot x\sin x$
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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,
$\frac{f'(c)g(c)-f(c)g'(c)}{[g(c)]^2}$

6. ## Quotient 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

7. 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)