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## Main Question or Discussion Point

Hello,

I'am suffering with the theoretical background. My course state the follow thing:

D(f-1(y))=1/D(f(x)).

So: f-1(y) is the inverse function of f(x), this means that the argument of f-1(y) is y!

Example: y = f(x) = x^2 => f-1(y): x = y^2. Am I correct with is one ?

The chain rule is applied at the following equation: D(f-1(y))=D(f-1(f(x)))=

(f-1(y))'.(f(x))'=1, this is because (f-1(f(x)) = x.

I do understand this, but I'm very confused with the next on:

(f-1(y))'=d(f-1(y))/dy, this is because the inverse function has y as argument.

Applied at my example I would get: d(f-1(y))/dy = 2.y

And by D(f-1(y))=1/D(f(x)), the derivative of the inverse function would be: 1/(d(x^2)/dx)

and this is 2.x

I TOTALLY don't understand it..

Someone dissolve my confussion ?

greetz

I'am suffering with the theoretical background. My course state the follow thing:

D(f-1(y))=1/D(f(x)).

So: f-1(y) is the inverse function of f(x), this means that the argument of f-1(y) is y!

Example: y = f(x) = x^2 => f-1(y): x = y^2. Am I correct with is one ?

The chain rule is applied at the following equation: D(f-1(y))=D(f-1(f(x)))=

(f-1(y))'.(f(x))'=1, this is because (f-1(f(x)) = x.

I do understand this, but I'm very confused with the next on:

(f-1(y))'=d(f-1(y))/dy, this is because the inverse function has y as argument.

Applied at my example I would get: d(f-1(y))/dy = 2.y

And by D(f-1(y))=1/D(f(x)), the derivative of the inverse function would be: 1/(d(x^2)/dx)

and this is 2.x

I TOTALLY don't understand it..

Someone dissolve my confussion ?

greetz