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:

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

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