Question

find dy/dx


y=cos^-1[(1-x^2)/(1+x^2)]

Answer 1

\[y = \cos ^{-1}(\frac{ 1-x ^{2} }{ 1+x ^{2} })\]

Answer 2

yes

Answer 3

step1; find the derivative of cos inverse

Answer 4

step2; use chain rule for the function inside, you would need to chain rule

Answer 5

and quotient rule

Answer 6

okay , what have told me , i have done all that
my answer coming is
2/(1+x^2)

Answer 7

so, is my anwer correct

Answer 8

where did cos inverse disappear? It will still be there

Answer 9

so first you find cos inverse derivative, then multiply it with the chain rule of the function inside cos inverse

Answer 10

your ans would be
derivative of cos inverse x chain rule

Answer 11

k
what i did is
i write cos y =(1-x^2)/(1+x^2)
then i solve it

Answer 12

cosy =( 1-tan^2y/2)/(1+tan^2y/2)
( 1-tan^2y/2)/(1+tan^2y/2)=( 1-x^2/2)/(1+x^2/2)

Answer 13

on comparing, we get,
tan(y/2)= x
then i differentiate wrt x

Answer 14

then i get
dy/dx= 2 /sec^2(y/2)
dy/dx=2/[1+tan^2(y/2)]
dy/dx=2/[1+x^2]

Answer 15

this is my answer

Answer 16

http://tutorial.math.lamar.edu/Classes/CalcI/DiffInvTrigFcns.aspx

Answer 17

warning - answer/solution - http://www.wolframalpha.com/input/?i=derivative+of+arccos%28%281-x%5E2%29%2F%281%2Bx%5E2%29%29

Answer 18

@stamp
what r u trying to say

Answer 19

@Kanwar245
plse tell me is my answer right

Answer 20

u mean \[\arccos *\frac{ 1-x^2 }{ 1+x^2 }\]

Answer 21

it is coming to be \[\frac{ (x^2+1)(-2x \cos^{-1}x -\sqrt{(1-x^2)}-2x(1-x^2)\cos^{-1} x }{ (x^2+1)^2 }\]