Derivatives of inverse functions-how two formulas relate?

[al_famky]

Derivatives of inverse functions--how two formulas relate?

Homework Statement

I know two formulas for calculating the derivative of an inverse function, both of which I know how to derive, but I don't know how to relate them to one another.

Homework Equations

$\frac{df^{-1}(y)}{dy}$=$\frac{1}{\frac{df(x)}{dx}}$
$f'^{-1}(x)$=$\frac{1}{f'(f^{-1}(x))}$

The Attempt at a Solution

Well, for the first formula, it's because
$\frac{dy}{dx}$=$\frac{df(x)}{dx}$
$\frac{dx}{dy}$=$\frac{df^{-1}(y)}{y}$
so
you get the first equation above.

for the second formula, use
$\left(f(f^{-1}(x))\right)$=$x$
take the derivative of x of both sides
$\left(f'\left(f^{-1}(x)\right)\right)$$\left(f'^{-1}(x)\right)$=1
and with a simple algebraic manipulation you get the second formula

so, that being said...
I don't think that the y's and x's of the two formulas mean the same thing.
if i was given an equation like
$y=arctan(-x^{3}$)
then how would I use the two formulas (without using $arctan'x=\frac{1}{1+x^{2}}$)

I started out with the first formula
stating $x=arctan(-y^{3}$), so $-y^{3}=tanx$, $y=-tan^{\frac{1}{3}}x$
$\frac{df^{-1}(y)}{dy}$=$\frac{1}{\frac{dy}{dx}}$=$\frac{sec^{2}}{-\frac{1}{3}tan^{-\frac{2}{3}}x}$
or something like that... the answer is easy enough to get, but is it the answer i want? I calculated the derivative on the left with respect to y, and on the right i get an equation with x...which means i have to somehow change y to x, and there is no really convenient way (this formula was supposed to make things more convenient for calculation you know.)

for the second formula...things are easier, because both sides calculate derivates with respect to x, but that's where I get confused.
Are the two formulas related in any way? They should, but I can't wrap my mind around which y and which x they represent.

I don't know if i presented my problem clearly enough, but I hope someone would be willing to clarify the matter for me, thank you for your time.

 

[mighty2000]

I can understand your confusion about how these two formulas for calculating the derivative of an inverse function are related. It can be a bit tricky at first, but with some careful analysis, we can see that they are indeed related.

First, let's clarify the variables used in these formulas. In the first formula, \frac{df^{-1}(y)}{dy} represents the derivative of the inverse function with respect to y. This means that y is the independent variable and x is the dependent variable. In the second formula, f'^{-1}(x) represents the inverse function itself, so x is the independent variable and y is the dependent variable.

Now, let's take a closer look at the two formulas:

\frac{df^{-1}(y)}{dy}=\frac{1}{\frac{df(x)}{dx}}

and

f'^{-1}(x)=\frac{1}{f'(f^{-1}(x))}

In the first formula, we are taking the derivative of the inverse function with respect to y, which means we are looking at the rate of change of y with respect to x. This is the same as saying we are looking at the slope of the tangent line to the inverse function at a point where the y-coordinate is y. This is why the derivative is written as \frac{df^{-1}(y)}{dy}.

In the second formula, we are taking the derivative of the inverse function itself, which means we are looking at the rate of change of x with respect to y. This is the same as saying we are looking at the slope of the tangent line to the original function at a point where the x-coordinate is x. This is why the derivative is written as f'^{-1}(x).

Now, if we use the chain rule to differentiate the second formula, we get:

f'^{-1}(x)=\frac{1}{f'(f^{-1}(x))}

f'^{-1}(x)f''(f^{-1}(x))=-\frac{f''(f^{-1}(x))}{f'(f^{-1}(x))^2}

f''(f^{-1}(x))=-\frac{f''(f^{-1}(x))}{f'(f^{-1}(x))^2}

Now, if we substitute this into the first formula, we get:

\frac{df^{-1}(y