# How do you perform inversions for #y = x^2 and y = x^4?# Is #(dx)/(dy)# from the inverse #1/((dy)/(dx))?#

##### 2 Answers

#### Explanation:

Assuming we are looking for the inverse functions of

and

we must consider a restricted domain. A function can only have an inverse function on a domain in which it is

Again, assuming we are only considering real-valued functions, we can partition

With that, we can now consider our restricted domains for

If we have

and

then we can find their inverse functions as

and

If we have

and

then we can find their inverse functions as

and

Note that in each of the above cases, we find that the needed equalities of

Graphically, we can also see why this is necessary. If

This is also why when solving

While we do not arrive there by multiplying by differentials in the typical manner for reciprocals, we can show that

Suppose

Disambiguation:

We are faithful to standard notations that involve

With prefix as a function operator and suffix as operand, we have

the angle

Here sin is a prefix function operator and x is the suffix operand.

If

Here again, f is function prefix and y is the operand.

Commutative law is not applicable to the chain operator

A MON AVIS: If y = f(x) and y is locally bijective ( in epsilon-

neighborhood ), the inverse relation is

For example,

if

with y cyclically

Now, x

In our problem here,

for

the inverse is defined piecewise as follows.

Likewise,

for

the inverse is defined as follows.

In either case, this piecewise x is differentiable everywhere for

x'=

it can also be verified that

It is important, that

the graphs of

So,, it is easy to see that the horizontal test and

vertical test are irrelevant.