IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - Differentiation of an Inverse Trigonometrical Function

nsv_j89

y=sin^-1(sinx)

dy/dx=? Please Explain.

A.Vignesh

y=sin^-1(sin x)

y = x [sin ^ -1(sin x) =x]

dy/dx = 1

KAB

A.Vignesh is right.

krishna.gopal

Sorry but here i would like to differ. For example sin^-1(sin(2.5

) is sin^-1(1) =

/2 and not 2.5

. So if y = sin^-1(sin(x))

then dy/dx is not defined at x=(2n+1)

for all other values it is 1