IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

rtiit

d/dx[sin^-1{cos(x²-1)}}=?

uday_zingtudor

simple one
using chain rule,

[1/

(1-cos²(x²-1))]*2cos(x²-1)sin(x²-1)*2x

that equals 4xcos(x²-1)

~Cheerio!!!

rtiit

i think der is some mistake
how did u get 2cos[x(square) -1] in the denominator???

ans is -2x

by d way i got it

sandeepramesh

note cos(x^2-1) = sin(pi/2-x^2 + 1) and that kills the problem :)

uday_zingtudor

d/dx{sin^_1cos(x^2-1)}=1/sin(x^2-1) 2cos(x^2-1)sin(x^2-1)2x

And see @rtiit 2cos(x^2-1) is in numerator not denominator

=2cos(x^2-1)2x

=4xcos(x^2-1)

Tell me how can u support ur answer

rtiit

see
dy/dx=1/[root{1-cos^2(x^2-1)}] *{-sin(x^2-1)}*2x

solving we get -2x

uday_zingtudor

okay sorry

sometimes my mind goes doing some rubbish

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