# Derivative of inverse trig function

VanKwisH

## 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??

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

VanKwisH
care to explain??

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

$$x=\cot y$$

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

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

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

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

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

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

Last edited:
VanKwisH
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

VanKwisH
so will anyone show me/explain how it works

rohanprabhu
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...

VanKwisH
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??

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

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

VanKwisH
did you not read my post?

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

thnx a lot

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.