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

1 Answer
Shwetank Mauria
Oct 1, 2016

Derivative of #y=sin^(-1)x# at #x=3/5# is #4/5#

Explanation:

As #y=sin^(-1)x#, we have

#x=siny# and #(dx)/(dy)=cosy#

and #(dy)/(dx)=1/cosy=1/sqrt(1-sin^2y)=1/sqrt(1-x^2)#

and derivative of function at #x=3/5# 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]}

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