Question

Can someone explain why this happens?

Say we have a function of `x`, for example `y=x^(1/2)`, we can also write this as `x=y^2`

By taking the derivative of `y=x^(1/2)`, we get `y'=x^(-1/2)/2`

By implicity differenciating `y^2=x` we get `y'=1/(2y)`

If we make `1/(2y)=x^(-1/2)/2` and rearrange to make `y` the subject, we get `y=x^(1/2)`, which was the original function.

However, for other functions, i.e. `y=sqrt(arcsin(x))` the function for equating both differentials gives a graph which include `y=sqrt(arcsin(x))` along with some other stuff.

Answer 1

You should get the same graph as the original function.

Explanation:

If `y = sqrt(arcsin(x))`

`{dy}/{dx} = 1/(2sqrt((1-x^2)arcsin(x)))`

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

`x = sin(y^2)`

Differentiating implicitly gives

`1 = cos(y^2) (2y) {dy}/{dx}`

or

`{dy}/{dx} = 1/(2ycos(y^2))`

If you want to equate the 2 "different" `{dy}/{dx}` to verify that they are the same, go ahead.

`1/(2sqrt((1-x^2)arcsin(x))) = 1/(2ycos(y^2))`

Simplifying it gives

`(1-x^2)arcsin(x) = y^2cos^2(y^2)`

which seem complicated but notice that the original equation `y=sqrt(arcsin(x))` is a solution (which should be the case)

Answer 2

I think I understand.

Explanation:

If I graph

`1/(2sqrt((1-x^2)arcsinx)) = 1/(2sqrtarcsinxcosy^2)`

I get:

The little blue piece is `y = sqrtarcsinx`

For `y = sqrtarcsinx`, the restrictions are `x in [0,1]` and `y in [0, sqrt(pi/2)]`

There is no such restriction on the `y` values for `1/(2sqrt((1-x^2)arcsinx)) = 1/(2sqrtarcsinxcosy^2)`.