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

1 Answer
mason m
Sep 19, 2016

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

Explanation:

Note that:

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

So, according to the chain rule:

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

So, where #f(x)=2x+1#, and #f'(x)=2#:

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

#=2/sqrt(1-(4x^2+4x+1))#

#=2/sqrt(-4x^2-4x)#

#=2/(sqrt4sqrt(-x^2-x))#

#=1/sqrt(-x^2-x)#

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