explanation of $ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} $?

Question

I'm studying about derivative of inverse function. The teacher in the video (https://www.youtube.com/watch?v=3ReOtNCYuBw) (at 9:00 minute) said this

if a differentiable function, f has an inverse, then:

$$ \frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'[f^{-1}(x)]} $$

provided $f'[f^{-1}(x)]\neq 0$

then he said if we make $y = f^{-1}(x)$ then:

$$ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} $$

the last line is when I really get lost because it should be $ \frac{dy}{dx} = \frac{1}{f'[y]} $ not $ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} $ isn't it? any please explain to me in very detail, I'm a newbie.

Answer 1

To say that $y = f^{-1}(x)$ is also to say that $f(y) = x$. Differentiating both sides with respect to $y$ yields $f'(y) = dx/dy$.

EDIT: This is only a matter of notation, and, unfortunately, there is some abuse of notation going on here. The first line you wrote is absolutely correct, whereas the statement $$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$$ is ambiguous, albeit easy to remember. One problem with the notation is that it is not clear what $x$ and $y$ represent (are they both functions? if so, what does it mean to differentiate with respect to a function?). Also, if the left-hand side is to be evaluated at a point $t$, then implicit is the fact that, on the right-hand side, $dx/dy$ should be evaluated at $y(t)$.

With this in mind, let us suppress $f$ from the notation. If $x$ is an invertible function of $y$, we may write $y$ as a function of $x$. Thus $x$ and $y$ act both as functions and coordinates, depending on the context. In words, your theorem says that "the derivative of the inverse of $x$ (i.e., the derivative of $y$) at a point $t$ is the inverse of the derivative of $x$ at the point $y(t)$". We can write this as

$$\frac{dy}{dx}(t) = \frac{1}{\frac{dx}{dy}(y(t))}.$$

If we suppress the points from the notation, we arrive at the statement above. So perhaps it would be more adequate to write

$$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy} \circ y}.$$

Answer 2

This short hand notation is best understood if you realize that you're identifying variables and functions with each other, namely $x,y$ and $f, f^{-1}$.

Let's digest the formula $$ \frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'[f^{-1}(x)]} $$ from the question. The left hand side is a function of $x$, namely the derivative of $g=f^{-1}$. It states that for any fixed value $t$ (read this as a specific value for $x$, we have that $g'(t) = \frac{d}{dx}[f^{-1}(x)](t)$ is given by the right hand side.

So the right hand side is a function of $x$, too. For $x=t$, its value is $\frac{1}{f'(f^{-1}(t))}$, or to put it in other terms: for $s =f^{-1}(t)$ or equivalently $t=f(s)$, we have that

$$\left(\frac{d}{dx}[f^{-1}(x)]\large\right)_{x=t} = \left(\frac{1}{f'(f^{-1}(x))}\right)_{x=t} = \left(\frac{1}{f'(y)}\right)_{y=s}.$$

Here, I used $y$ as the name of the variable of $f$ and $f'$. We can write $f'(y)=\frac{d}{dy}[f(y)]$, so that the equation above becomes:

$$\left(\frac{d}{dx}[f^{-1}(x)]\large\right)_{x=t} = \left(\frac{1}{\frac{d}{dy}[f'(y)]}\right)_{y=s}.$$

In this equation, $x$ and $y$ still denote variables, not functions. If we keep in mind that for the fixed values $s,t$ above we have $f(s)=t$ and $f^{-1}(t)=s$, we can write for the variables $x,y$: $$y=f^{-1}(x) \mbox{ and } x=f(y).$$ Now, if we view these equations as definitions of functions named $x$ and $y$, we end up with:

$$\left(\frac{d}{dx}[y(x)]\right)_{x=t} = \left(\frac{1}{\frac{d}{dy}[x(y)]}\right)_{y=s}.$$

Lose the dependencies from the variables $x,y$ and their respective values and you end up with $$\frac{d}{dx}[y]= \frac{1}{\frac{d}{dy}[x]}.$$

Answer 3

I repeat the comments as they are getting too long.

There are two notations for the derivative of a function: $$F'(X) = \frac{d F}{dX}$$ On the right hand side, the $F$ in the numerator should be a function and in the denominator the $X$ should be the variable of the function. (To quibble more, there is a small mistake, if one really wants an equality of value, i.e. when a function is evaluated at a point, then one should write $F'(X_0) = \frac{d F}{dX}(X_0)$, otherwise it could be an equality of functions $F' = \frac{d F}{dX}$).

Assuming the first equality is prooved, the second $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$ is just a matter of notations, it is a rewritting of the first.

• in the left hand side (l.h.s.), one needs to understand $y$ as a function of $x$, namely $y: x\rightarrow f^{-1}(x)$ or in short $y=f^{-1}(x) $
• in the r.h.s., $x$ has to be understood as a function $x: y \rightarrow f(y)$ of $y$ but to avoid all possible confusion one should actually write $X$ for this function and $Y$ its variable since $x$ was previously already used as a variable and $y$ as a function.

Now where is everthing evaluated at, and to begin with, what is the variable (which should be the same in both side): start looking at l.h.s., $y$ is a function of $x$ and should be evaluated at some $x_0$. R.h.s. is then evaluated at $Y_0 := y(x_0)$ in order to recover the first equality so one could write $$\frac{dy}{dx}(x_0) = \frac{1}{\frac{dX}{dY}(Y_0)}$$

Answer 4

A quick visualisation of this which I've always enjoyed is that if we draw the tangent to $y=f(x)$ at $(a,b)$, then its equation will be of the form $\frac{y-b}{x-a}=m$.

It thus follows naturally that if we draw the tangent to $x=f(y)$ at $(b,a)$, then its equation will be of the form $\frac{x-a}{y-b}=\frac{1}{m}$.

Drawing together the point-slope form of straight lines and the interpretation of $\frac{dy}{dx}$ as the slope gives quite neatly that $\frac{dy}{dx}\frac{dx}{dy}=1$