How do you find the derivative of $y=sin^-1(2x+1)$ ?

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Answer 1 · 2018-04-18T16:52:32

The answer is $y'=(2)/sqrt(1-(2x+1)^2$

$d/dx[y]=[1/sqrt(1-(2x+1)^2]*d/dx[2x+1]]$

$y'=(2)/sqrt(1-(2x+1)^2$

How do you find the derivative of y=sin^-1(2x+1)y=sin^-1(2x+1)?