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

1 Answer

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

Explanation:

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}}