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

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Answer 1 · 2016-10-01T09:48:58

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

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]}

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