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

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Answer 1 · 2016-10-28T08:01:21

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

$y=sin^-1(2x+1)$

$=>2x+1=siny$

$:.2(dx)/dy=cosy$

$=>(dy)/dx=2/cosy$

using $cos^2y+sin^2y=1$

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

substitue back for x

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

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