What is the derivative of arcsin sqrt(2x)?

Jun 6, 2015

The derivative is $\frac{1}{\setminus \sqrt{2 x} \setminus \sqrt{1 - 2 x}}$

Let $y = \arcsin \setminus \sqrt{2 x}$

The goal is to solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$. Take $\sin$ of each side of the bove equation and get,

$\sin \left(y\right) = \setminus \sqrt{2 x}$

Take the derivative of each side with respect to $x$

$\cos \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\setminus} \sqrt{2 x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\setminus \sqrt{2 x} \cos \left(y\right)}$

We need to know what $\cos \left(y\right)$ is. Use $\sin \left(y\right) = \setminus \frac{\sqrt{2 x}}{1} = \setminus \frac{\textrm{o p p o s i t e}}{\setminus} \textrm{h y p o t e \nu s e}$ to get the triangle below.

From the diagram $\cos \left(y\right) = \frac{\setminus \sqrt{1 - 2 x}}{1}$. Substitute this into the $\frac{\mathrm{dy}}{\mathrm{dx}}$ expression to get,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\setminus \sqrt{2 x} \setminus \sqrt{1 - 2 x}}$

Alternatively, use the chain rule.

Given,

$\frac{d}{\mathrm{dz}} \left[\arcsin \left(z\right)\right] = \frac{1}{\setminus} \sqrt{1 - {z}^{2}}$

Let $z = \setminus \sqrt{2 x}$

$\frac{d}{\mathrm{dx}} \left[\arcsin \left(\setminus \sqrt{2 x}\right)\right] = \frac{d}{\mathrm{dz}} \left[\arcsin \left(z\right)\right] \frac{\mathrm{dz}}{\mathrm{dx}}$

Substitute $z = \setminus \sqrt{2 x}$ to get the same answer as before

$\frac{d}{\mathrm{dx}} \left[\arcsin \left(\setminus \sqrt{2 x}\right)\right] = \frac{1}{\setminus \sqrt{2 x} \setminus \sqrt{1 - 2 x}}$