How do you find the derivative of the function: y=(sin^-1x)y=(sin1x), at x=3/5?

1 Answer
Oct 1, 2016

Derivative of y=sin^(-1)xy=sin1x at x=3/5x=35 is 4/545

Explanation:

As y=sin^(-1)xy=sin1x, we have

x=sinyx=siny and (dx)/(dy)=cosydxdy=cosy

and (dy)/(dx)=1/cosy=1/sqrt(1-sin^2y)=1/sqrt(1-x^2)dydx=1cosy=11sin2y=11x2

and derivative of function at x=3/5x=35 is

y'(3/5)=1/sqrt(1-(3/5)^2)=1/sqrt(1-9/25)=1/sqrt(16/25)=1/(4/5)=5/4
graph{arcsinx [-5, 5, -2.5, 2.5]}