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

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Answer 1 · 2016-12-30T14:23:17

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

you can expand the brackets and simply if you wish.In which case you get:

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

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

$=>2x+1=siny$

differentiate wrt $x$

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

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

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

$=>(dy)/(dx)=2/sqrt(1-(4x^2+4x+1))=2/sqrt(-4x(x+1))$

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

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