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

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Answer 1 · 2016-05-21T01:15:27

#d/dxarcsin(2x+1) = 2/(sqrt(1-(2x+1)^2)#

A useful trick for deriving the derivatives of inverse trig functions is to use implicit differentiation:

Let #y = arcsin(2x+1)#

#=> sin(y) = 2x+1#

#=> d/dxsin(y) = d/dx(2x+1)#

#=> cos(y)dy/dx = 2#

#=> dy/dx = 2/cos(y)#

Drawing a right triangle with an angle #y# such that #sin(y) = 2x+1#, we find that #cos(y) = sqrt(1-(2x+1)^2)#. Thus we get our final result:

#d/dxarcsin(2x+1) = dy/dx = 2/(sqrt(1-(2x+1)^2)#