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))#