# What is the derivative of arcsin(2x+1)?

Aug 29, 2015

${y}^{'} = \frac{1}{\sqrt{- x \left(x + 1\right)}}$

#### Explanation:

You can differentiate this function by using implicit differentiation. Start from

$y = \arcsin \left(2 x + 1\right)$

This means that you have

$\sin y = 2 x + 1$

Now differentiate both sides with respect to $x$

$\frac{d}{\mathrm{dx}} \left(\sin y\right) = \frac{d}{\mathrm{dx}} \left(2 x + 1\right)$

$\cos y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 2$

Isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$ on one side of the equation

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{\cos} y$

Use the trigonometric identity

$\textcolor{b l u e}{{\cos}^{2} y + {\sin}^{2} y = 1}$

to write $\cos y$ as a function of $\sin y$

${\cos}^{2} y = 1 - {\sin}^{2} y$

$\sqrt{{\cos}^{2} y} = \sqrt{1 - {\sin}^{2} y}$

$\cos y = \sqrt{1 - {\sin}^{2} y}$

This means that you have

(dy)/dx = 2/sqrt(1-sin^2y) = 2/(sqrt(1 - (2x+1)^2)

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{\sqrt{\textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} - 4 {x}^{2} - 4 x - \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}}}} = \frac{2}{\sqrt{4 x \left(- x - 1\right)}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \sqrt{- x \left(x + 1\right)}} = \textcolor{g r e e n}{\frac{1}{\sqrt{- x \left(x + 1\right)}}}$