Differentiate Inverse Sine (arcsin) - proof

How to differentiate cos^-1x

y=cos^-1x

Bring the cos across

cosy = x

Differentiate both sides, remember when differentiating y time by dy/dx

-sin(y) dy/dx = 1

dy/dx = -1/siny

However we want to get the differential in terms of x, to do this we can use the identity

sin²t+cos²t = 1

so

sint = √(1 - cos²t)

putting this into our expression for dy/dx we get

dy/dx = 1/√(1-cos²y)

but cosy = x so

dy/dx = 1/√(1-x²)

by David Woodford

Differentiate Inverse Sine

This tutorial explain how to differentiate inverse sine, this applies when using radians.

begin with

y = sin^-1 x
bring sin^-1 across to become sin
sin y = x
differentiate
cos y dy/dx = 1
note that the derivative of sint wrtt is cos t as explained in an earlier tutorial and by the chain rule when we differentiated sin y it became cosy time dy/dx as we are differnetiatiny a y and the derivative of y is dy/dx

then make dy/dx the subject

dy/dx = 1/cosy

We know the identity
sin²t + cos²t = 1
so we can wrtie
cos t =√(1 - sin²t)

we can now put this into the expression for dy/dx to get
dy/dx = 1/√(1 - sin²y)
but we know from the second line that sin y = x so

dy/dx = 1/√(1 - x²)