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

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Answer 1 · 2018-07-28T01:26:55

#f'(x)=\frac{2}{\sqrt{1-(2x+1)^2}}#

Given function:

#y=\sin^{-1}(2x+1)#

differentiating above function w.r.t. #x# using chain rule as follows

#d/dxf(x)=d/dx\sin^{-1}(2x+1)#

#=\frac{1}{\sqrt{1-(2x+1)^2}}d/dx(2x+1)#

#=\frac{1}{\sqrt{1-(2x+1)^2}}(2)#

#=\frac{2}{\sqrt{1-(2x+1)^2}}#