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

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Answer 1 · 2016-10-26T22:53:47

$dy/dx=1/sqrt(-x^2-x)$

rewrite $y=arcsin(2x+1)$ as $siny=2x+1$

Differentiating implicitly we get;
$cosydy/dx=2$
$:. dy/dx=2/cosy$

Using $sin^2A+cos^2A-=1 => (2x+1)^2+cos^2y=1$
$:. 4x^2+4x+1+cos^2y=1$
$:. cos^2y=-4x^2-4x$
$:. cos^2y=4(-x^2-x)$
$cosy=2sqrt(-x^2-x)$

and so $:. dy/dx=2/cosy => dy/dx=2/(2sqrt(-x^2-x))$
Hence, $dy/dx=1/sqrt(-x^2-x)$

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