# Derivative of inverse trig function

## Homework Statement

Is there a derivative of cot^-1x

## The Attempt at a Solution

so far i used the inverse property to detemermine that
coty = x
then i used the chain rule to get
-csc^2y * dy/dx = 1
and i isolated for dy/dx
dy/dx = 1 / -csc^2y

what now??

mjsd
Homework Helper
$$-\frac{1}{x^2+1}$$

care to explain??

$$y=\cot^{-1}x$$

$$x=\cot y$$

$$\frac{d}{dy}(x)=\frac{d}{dy}(\cot y)$$

$$1=-\csc^2 yy'$$

$$y'=-\frac{1}{\csc^2 y}$$

$$*1+\cot^2 y=\csc^2 y*$$

$$*x^2=\cot^2 y*$$

$$y'=-\frac{1}{x^2 +1}$$

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yah hahaha i'm lookin at the things in the tex editor and i dono't know wtf is goin on but i'll learn sooon :P

so will anyone show me/explain how it works

care to explain??

This is a standard integral. By 'standard' i mean.. this is one of the integrals you are just supposed to know. However, it is proved using Integration by parts. Almost all books on calculus carry that proof...

This is a standard integral. By 'standard' i mean.. this is one of the integrals you are just supposed to know. However, it is proved using Integration by parts. Almost all books on calculus carry that proof...

EDIT:
my calculus book says it's
$$\pi / 2-\tan^-1x$$

But i have no idea how to prove it anyone know??

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or can anyone walk me through this .......
i'm at $$\frac{d}{dx}$$ = $$\frac{1}{/-csc^2x}$$

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or can anyone walk me through this .......
i'm at $$\frac{d}{dx}$$ = $$\frac{1}{/-csc^2x}$$
did you not read my post?

did you not read my post?

ahhh srry i only saw the first half where u said
"still typing"

thnx a lot

Gib Z
Homework Helper
EDIT:
my calculus book says it's
$$\pi / 2-\tan^-1x$$

But i have no idea how to prove it anyone know??

What are u referring to? The integral, derivative, or the inverse cotangent itself can guess it, but you should make it clear.

It would be good if in the next post you just re cap the things you need help on.