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

Assuming we are looking for the inverse functions of #f(x) = x^2# and #g(x) = x^4#, i.e., #f^(-1)(x)# and #g^(-1)(x)# such that

#f(f^(-1)(x))=f^(-1)(f(x)) = x#
and
#g(g^-1(x)) = g^-1(g(x)) = x#

we must consider a restricted domain. A function can only have an inverse function on a domain in which it is #1-1#, or else there would be ambiguity as to what to map points in the codomain back to. For example, if our initial domain is #RR#, should we have #f^-1(4)=2# or #f^(-1)(4)=-2#?

Again, assuming we are only considering real-valued functions, we can partition #RR# into two domains, each of which results in #x^2# and #x^4# being #1-1#. We will use the notation that for #x>0#, #sqrt(x)# is the principal square root of #x#, that is, the unique value #y in RR^+# such that #y^2=x#. Similarly, #root(4)(x)# will represent the unique #y in RR^+# such that #y^4 = x#.

With that, we can now consider our restricted domains for #x^2# and #x^4#.

If we have
#f(x):[0, oo)->[0, oo), f(x) = x^2#
and
#g(x):[0, oo)->[0, oo), g(x) = x^4#

then we can find their inverse functions as

#f^-1(x):[0, oo)->[0, oo), f^-1(x) = sqrt(x)#
and
#g^-1(x):[0, oo)->[0, oo), g^-1(x) = root(4)(x)#

If we have
#f(x):(-oo,0]->[0, oo), f(x) = x^2#
and
#g(x):(-oo,0]->[0, oo), g(x) = x^4#

then we can find their inverse functions as

#f^-1(x):[0, oo)->(-oo,0], f^-1(x) = -sqrt(x)#
and
#g^-1(x):[0, oo)->(-oo,0], g^-1(x) = -root(4)(x)#

Note that in each of the above cases, we find that the needed equalities of #f(f^(-1)(x))=f^(-1)(f(x)) = x# and #g(g^-1(x)) = g^-1(g(x)) = x# are satisfied without ambiguity due to the restricted domains.

Graphically, we can also see why this is necessary. If #(x_0, y_0)# is a point on the graph of #y=f(x)#, then #(y_0, x_0)# should be a point on the graph of #y = f^-1(x)#. We can find the graph of #y=f^-1(x)# by reflecting the graph #y=f(x)# about the line #y=x#. If #y=f(x)# is not #1-1#, however, it will not pass the horizontal line test, meaning #f^-1(x)# will not pass the vertical line test, and thus will not be a function.

This is also why when solving #x^2=a#, we get #x = +-sqrt(a)#, as there is no non-multivalued function which is the inverse of #x^2# on #RR#.

While we do not arrive there by multiplying by differentials in the typical manner for reciprocals, we can show that #dx/dy = 1/(dy/dx)# via the chain rule .

Suppose #y = f(x)#. Differentiating with respect to #y#, we get

#d/dyy = d/dyf(x)#

#=> 1 = dx/dy (df(x))/dx# (chain rule)

#=> 1 = dx/dy dy/dx#

#:. dx/dy = 1/(dy/dx)#

Disambiguation:

We are faithful to standard notations that involve #..()^(-1)# ,

#x^(-1)=1/x.# Here, the operand is a prefix. This a particular case of

#x^(-n)=1/x^n#.

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

#sin^(-1) x# restricted to the principal value of

the angle # in [-pi/2, pi/2]#, whose sine is x.

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

If #y = f(x)#,

#x = f^(-1)(y)#

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

Commutative law is not applicable to the chain operator #f f^(-1)#.

#ff^(-1)# is not equivalent to #f^(-1)f#.

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

neighborhood ), the inverse relation is

#x = f^(-1)(y)#.

For example,

if #y = sin x in [-1, 1]#, I define x piecewise as

# f^(-1)(y) = x = kpi+(-1)sin^(-1)y#, for

#k = 0, +-1, +-2, +-3, ..., #,

with y cyclically #in [-1, 1]#.

Now, x #rarrlarry#, making y a locally bijective function.

In our problem here,

for # y = x^2 in [0, oo), x in ( -oo, oo )#,

the inverse is defined piecewise as follows.

#f^(-1)(y)=x#

#= -sqrt y, x in (-oo, 0] #

#= sqrt y, x in [0, oo)#.

Likewise,

for #y = x^4, in [0, oo), x in ( -oo, oo )#,

the inverse is defined as follows.

#f^(-1)(y)=x#

#= -sqrt sqrt y, x in (-oo, 0] #

#= sqrt sqrt y, x in [0, oo)#.

In either case, this piecewise x is differentiable everywhere for

x'=#(dx)/(dy)# and

it can also be verified that #x'=1/(y')#.

It is important, that

the graphs of # y = f(x) and x = f^(-1)(y)# are one and the same.

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

vertical test are irrelevant.