# Can someone explain why this happens?

## Say we have a function of $x$, for example $y = {x}^{\frac{1}{2}}$, we can also write this as $x = {y}^{2}$ By taking the derivative of $y = {x}^{\frac{1}{2}}$, we get $y ' = {x}^{- \frac{1}{2}} / 2$ By implicity differenciating ${y}^{2} = x$ we get $y ' = \frac{1}{2 y}$ If we make $\frac{1}{2 y} = {x}^{- \frac{1}{2}} / 2$ and rearrange to make $y$ the subject, we get $y = {x}^{\frac{1}{2}}$, which was the original function. However, for other functions, i.e. $y = \sqrt{\arcsin \left(x\right)}$ the function for equating both differentials gives a graph which include $y = \sqrt{\arcsin \left(x\right)}$ along with some other stuff.

Feb 19, 2018

You should get the same graph as the original function.

#### Explanation:

If $y = \sqrt{\arcsin \left(x\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{\left(1 - {x}^{2}\right) \arcsin \left(x\right)}}$

On the other hand, we can write the first equation as

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

Differentiating implicitly gives

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

or

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

If you want to equate the 2 "different" $\frac{\mathrm{dy}}{\mathrm{dx}}$ to verify that they are the same, go ahead.

$\frac{1}{2 \sqrt{\left(1 - {x}^{2}\right) \arcsin \left(x\right)}} = \frac{1}{2 y \cos \left({y}^{2}\right)}$

Simplifying it gives

$\left(1 - {x}^{2}\right) \arcsin \left(x\right) = {y}^{2} {\cos}^{2} \left({y}^{2}\right)$

which seem complicated but notice that the original equation $y = \sqrt{\arcsin \left(x\right)}$ is a solution (which should be the case)

Feb 19, 2018

I think I understand.

#### Explanation:

If I graph

$\frac{1}{2 \sqrt{\left(1 - {x}^{2}\right) \arcsin x}} = \frac{1}{2 \sqrt{\arcsin} x \cos {y}^{2}}$

I get:

The little blue piece is $y = \sqrt{\arcsin} x$

For $y = \sqrt{\arcsin} x$, the restrictions are $x \in \left[0 , 1\right]$ and $y \in \left[0 , \sqrt{\frac{\pi}{2}}\right]$

There is no such restriction on the $y$ values for $\frac{1}{2 \sqrt{\left(1 - {x}^{2}\right) \arcsin x}} = \frac{1}{2 \sqrt{\arcsin} x \cos {y}^{2}}$.