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))#