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

1 Answer
marfre
Feb 17, 2017

#1/sqrt(-x(x+1))#

Explanation:

Use the formula: #d/dx arcsin (u/a) = 1/sqrt(a^2-u^2) (du)/dx#

From the equation #y = arcsin(2x+1)#:

#a = 1# and #u = 2x+1# and #(du)/dx = 2#

so #y' = 2/sqrt(1^2-(2x+1)^2)#

Simplify the square root function:
#1^2-(2x+1)^2 = 1-(4x^2+4x+1) = -4x^2-4x #
#= -4x(x+1)#

so #y' = 2/sqrt(-4x(x+1)) = 2/(2sqrt(-x(x+1)) )= 1/sqrt(-x(x+1))#

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