Thread: Differentiate cosecx

Prove that d/dx (cosecx) = -cosecxcotx

Originally Posted by nath_quam

Prove that d/dx (cosecx) = -cosecxcotx

Hint: start with this
cscx = 1/sinx

-Dan

What would you like us to do with the attachment?

RonL

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

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,
$\displaystyle \frac{f'(c)g(c)-f(c)g'(c)}{[g(c)]^2}$

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

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)