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$$\frac{dy}{dx}= \frac{1}{\;\frac{dx}{dy}\;}$$ Why is the above theorem true as long as $dx/dy$ is not zero? How can you prove it rigorously?

I don’t think it is obvious by the definition of the derivative. I think this says $dx/d(x^2)$ will equal to $1/2x$ and so we can evaluate derivatives such as this. But I want a rigorous proof.

Edit: by the answers I think you want the existence and differentiability of f inverse for something like this to even work ? Could the derivative still exist in such an example and fail to be able to be evaluated like this?Or does that have no meaning ?

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    $\begingroup$ This is called the inverse function theorem. The confusing thing about it is that the evaluation points on the left and right sides are different. For example, this theorem says that the derivative of $x^2$ at $x=2$ is the reciprocal of the derivative of $\sqrt{y}$ at $y=4$. $\endgroup$ – Ian Jun 19 '20 at 16:11
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    $\begingroup$ If you want a rigorous proof, we need a rigorous formulation of the question, not some playing with symbols you did or did not understand.. $\endgroup$ – user436658 Jun 19 '20 at 16:13
  • $\begingroup$ The idea is that in the left member, you consider $y$ as a function of $x$, while in the right member, you consider $x$ as a function of $y$. For this to be possible, the function $f$ such that $y = f(x)$ has to be bijective. By definition $\frac{dy}{dx} = f'(x)$ and $\frac{dx}{dy} = g'(y)$, where $g$ is the inverse function of $f$. $\endgroup$ – DodoDuQuercy Jun 19 '20 at 16:13
  • $\begingroup$ @Ian I don’t understand I just want to know how to evaluate derivatives like dx/d(x^2) intuitively it should be 1/2x but how do we show it formally? $\endgroup$ – Pax Daga Jun 19 '20 at 16:13
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    $\begingroup$ What's throwing you off is that you're not keeping track of what is a function of what. For instance if $y=x^2$ when $x \geq 0$ then $x=\sqrt{y}$, so $\frac{dx}{dy}$ is a function of $y$. That function is actually $\frac{1}{2x}$ but $x$ is now a function of $y$, too, namely $x=\sqrt{y}$. So you've recovered a result you probably already knew about the derivative of the square root function. $\endgroup$ – Ian Jun 19 '20 at 16:16
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If $$ f^{-1}(f(x)) = x $$ in some neighborhood of $x$, then by the chain rule, $$ \dfrac{df^{-1}(f(x))}{dy} f'(x)= 1, $$ and $$ \dfrac{df^{-1}(y)}{dy}= \dfrac{1}{f'(x)} $$ where $y = f(x)$.

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  • $\begingroup$ Definitely assuming the existence and differentiability of $f^{-1}$. Also notice the "in a nbhd of $x$" comment: if $f(x) = x^2$, then the "inverse" on $\mathbb{R}$ is a correspondence and not even a function, $f^{-1}(y) = \{\sqrt{y},-\sqrt{y}\}$, for example, despite the InvFT being true in a small enough nbhd of any particular $x \in \mathbb{R}$. $\endgroup$ – user762914 Jun 19 '20 at 17:27
  • $\begingroup$ Not to complicate things too much, but in terms of "proof", the inverse function theorem and the implicit function theorem are related ideas and very general. The standard proof is to construct a contraction mapping for which the implicit function is a fixed point, and then use Banach's fpt to establish existence. Once you know how that works, you can relax some of the assumptions. There are versions of the Implicit Function Theorem that drop differentiability, replacing with a local injective assumption, and the result is a Lipschitz implicit function instead of a differentiable one. $\endgroup$ – user762914 Jun 19 '20 at 17:35
  • $\begingroup$ Take $x=2$, $y=4=x^2$. The function $\sqrt{x}:\mathbb{R} \rightarrow \mathbb{R}$ would have two values in the inverse, $2$ and $-2$. But the function $x^2 : (1,3) \rightarrow \mathbb{R}$ has a unique inverse, where $(1,3)$ is a nbhd of $x=2$. The $f^{-1}$ is only going to be well-defined globally if $f$ is injective and differentiable. Like anything with calculus, this is a local result that will only extend to the entire domain if the conditions are right. Think about $\sin$: $\sin^{-1}$ has countably many solutions for any $y \in [-1,1]$; at which one are you computing your inverse? $\endgroup$ – user762914 Jun 19 '20 at 17:47
  • $\begingroup$ Ok so then do you have to prove it for x>=0 and x<0 separately for $x^2 $ case ? $\endgroup$ – Pax Daga Jun 19 '20 at 18:00
  • $\begingroup$ Sure, $y = |x|$. The derivative does not exist at zero, so you cannot compute $f'(x)$, so the formula $D_y f^{-1}(0) = 1/f'(0)$ does not exist. It is worse than that, because at $0$, any nbhd of $0$, $|x|=y$ is inversely mapped to a set $\{x,-x\}$, and you cannot be sure if $f'(x)$ is $1$ or $-1$. Calculus is just a very local subject but you sound like you want a global answer, and there isn't one unless you want to add more assumptions that apply to your specific application. $\endgroup$ – user762914 Jun 20 '20 at 14:32
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This question has some good answers already, but I want to point out that the intuition from abusing the notation can lead to a proof directly.

Just using the limits behind the derivative notation works: $$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta y \to 0} \frac{\Delta y}{\Delta x} =\lim_{\Delta y \to 0} \frac{1}{\Delta x / \Delta y} = \frac{1}{\lim_{\Delta y \to 0}\Delta x / \Delta y} = \frac{1}{dx/dy}$$


But this does require some extra explaination. First of all, we assume $y = f(x), x = f^{-1}(y)$, i.e. $y$ is a function of $x$ and vice versa. We need that these functions are differentiable so that all limits written above exist. Next $$\Delta x = x_2 - x_1,\quad \Delta y = f(x_2) - f(x_1) = y_2 - y_1$$ or equivalently $$\Delta y = y_2 - y_1,\quad \Delta x = f^{-1}(y_2) - f^{-1}(y_1) = x_2 - x_1$$

Last but not least, because $f$ and $f^{-1}$ are continuous (because they are differentiable), we have that $$\Delta x \to 0\iff\Delta y \to 0$$ which I used in the beginning.

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  • $\begingroup$ If f inverse is not differentiable then this fails is there an example of such $\endgroup$ – Pax Daga Jun 20 '20 at 6:02
  • $\begingroup$ Please answer my edit $\endgroup$ – Pax Daga Jun 20 '20 at 6:36
  • $\begingroup$ @VivaanDaga What do you mean an example? If $f^{-1}$ is not differentiable then the theorem statement does not make sense. It contains the derivative of $f^{-1}$: $\frac{1}{dx/dy} = 1/(f^{-1})'(y)$ by definition. $\endgroup$ – Jens Renders Jun 21 '20 at 9:19
  • $\begingroup$ the theorem will fail but can the derivative still exist? i would not think so because you need to express x as a function of x in dx/df ? am i correct? $\endgroup$ – Pax Daga Jun 21 '20 at 9:34
  • $\begingroup$ @VivaanDaga Can what derivative still exist? $\endgroup$ – Jens Renders Jun 21 '20 at 9:48
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Let $y=f(x)$ with the inverse function of $x=g(y)$.

We have $$f(g(y))=y$$

Apply the chain rule to get$$ f'(g(y))g'(y) =1$$ Thus $$ f'(g(y))=\frac {1}{g'(y)}$$

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

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  • $\begingroup$ In this proof do you assume existence and differentiability of the inverse? $\endgroup$ – Pax Daga Jun 19 '20 at 17:25
  • $\begingroup$ @VivaanDaga It actually follows from the assumption that $f' \neq 0$, but this requires a little effort to prove rigorously. As written here the computation is "formal" which basically means "if all these symbols mean anything, the only thing they could possibly mean is this, but I haven't yet checked whether they make sense". $\endgroup$ – Ian Jun 19 '20 at 18:09
  • $\begingroup$ @Ian So do you have to different functions for x>=0 and for x<0 for the x^2 case because otherwise the inverse does not exist am i right or wrong? $\endgroup$ – Pax Daga Jun 19 '20 at 18:13
  • $\begingroup$ @VivaanDaga Yes, you have two different functions given by $f(x)=x^2$ that are invertible and differentiable: one is defined on $(0,\infty)$, the other is defined on $(-\infty,0)$. On $\mathbb{R}$, $f(x)=x^2$ is not invertible. Notably one can extend the domain of either one of these branches to include $0$, but if you do then the inverse isn't differentiable everywhere. This is no contradiction since $f'(0)=0$. $\endgroup$ – Ian Jun 19 '20 at 18:16
  • $\begingroup$ @Ian what happens to the derivatives? $\endgroup$ – Pax Daga Jun 19 '20 at 18:18
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Consider the chain rule.

But first off note, neither $\frac {dy}{dx}$ nor $\frac {dx}{dy}$ need not exist or make any sense. But if they do there there exist some function $f$ where:

$y = f(x)$ and $f$ is differentiable and $f$ is invertable so $x = f^{-1}(y)$ and $f^{-1}$ is differentiable. And we have that $x = f^{-1}(f(x))$ and $y = f(f^{-1}(y))$.

If we accept that is our premise we can just use the chain rule.

On the one hand we have the identity function $i(x) = x$ and $i'(x) =1$ or in Leibniz notation $\frac {dx}{dx} = 1$.

But if we view $i(x)$ as a composite function $i(x) = f^{-1}(f(x))$ then we can derive the derivative via the chain rule: we have $i'(x)=[f^{-1}]'(f(x))\cdot f'(x)=[f^{-1}]'(y)\cdot f'(x)$ or in Leibniz notation $\frac {dx}{dx} =\frac {d(f^{-1}(f(x))}{dx} = \frac {d(f^{-1}(f(x))}{d(f(x))}\frac {d(f(x))}{dx}=\frac {dx}{dy}\frac {dy}{dx}$

But bearing in mind that $i'(x) = 1$ or $\frac {dx}{dx} = 1$ we just manipulate:

$i'(x)=[f^{-1}]'(f(x))\cdot f'(x)=[f^{-1}]'(y)\cdot f'(x)=1$ so $f'(x)=\frac 1{[f^{-1}]'(f(x))}= \frac 1{[f^{-1}]'(y)}$. Or in Leibniz notation $\frac {dx}{dx} =\frac {d(f^{-1}(f(x))}{dx} = \frac {d(f^{-1}(f(x))}{d(f(x))}\frac {d(f(x))}{dx}=\frac {dx}{dy}\frac {dy}{dx}=1$ so $\frac {dy}{dx} = \frac 1{\frac {dx}{dy}}$.

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We could prove this directly similar to how we prove the chain rule:

$1 = \lim_{x\to a} \frac {x-a}{x-a} =\lim_{x\to a}\frac {f^{-1}(f(x))- f^{-1}(f(a))}{x-a}=$

$\lim_{x\to a} \frac {f^{-1}(f(x))- f^{-1}(f(a))}{f(x) - f(a)}\frac {f(x)-f(a)}{x-a}=$ (assuming $f$ is continuous and ... nice)

$\lim_{f(x)\to f(a)}\frac {f^{-1}(f(x))- f^{-1}(f(a))}{f(x) - f(a)}\lim_{x\to a} \frac {f(x)-f(a)}{x-a}=$

$[f^{-1}]'(f(a))\cdot f'(a)$.

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I think this says dx/d(x2) will equal to 1/2x and so we can evaluate derivatives such as this. But I want a rigorous proof.

If $y= x^2$ then $x =\begin{cases}\sqrt y&x\ge0\\-\sqrt y& x \le 0\end{cases}$

If $x \ge 0$ then $\frac {dx}{d(x^2)} = \frac {d\sqrt{y}}{dy}=\frac 1{2\sqrt{y}} = \frac 1{2x}$.

If $x \le 0$ then $\frac {dx}{d(x^2)} = \frac {d(-\sqrt{y})}{dy}= -\frac 1{2\sqrt{y}} = \frac 1{2x}$.

So $\frac {dx}{d(x^2)} = \frac 1{2x}$.

......

Provided there is an $f$ so that $y = f(x)$ and $x = f^{-1}(y)$. We can always have

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

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  • $\begingroup$ If the inverse is not differentiable this evaluation method will fail is that correct? also might the derivative still exist if the inverse is not differentiable? $\endgroup$ – Pax Daga Jun 19 '20 at 20:15
  • $\begingroup$ If the inverse is not differentiable then $\frac {dy}{dx}$ is meaningless. "also might the derivative still exist if the inverse is not differentiable?" Yes, your exapme of $y =x^2$ is a good example of that $x=\sqrt{y}$ is not differentiable at $x = 0$. $\endgroup$ – fleablood Jun 19 '20 at 20:22
  • $\begingroup$ If f inverse is not differentiable then this fails. is there an example of such $\endgroup$ – Pax Daga Jun 20 '20 at 6:02
  • $\begingroup$ Please answer my edit $\endgroup$ – Pax Daga Jun 20 '20 at 6:35
  • $\begingroup$ I already did. $x=\sqrt{y}$ is not differentiable at $x = 0$. $\endgroup$ – fleablood Jun 20 '20 at 7:39
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It can be shown quite easily that if $f:[a,b]\rightarrow [c,d] $ is differentiable some $x_0 \in [a,b]$ then there’s a function $\Phi $ that is continuous at $x_0 $ and $\Phi (x_0)=f’(x_0) $ and $$f(x)=f(x_0)+ \Phi (x)(x-x_0) $$. The converse also holds.

Suppose we have a continuous bijection $f:[a,b]\rightarrow [c,d] $ is differentiable at $x_0$ and $f’(x_0)\neq 0 $. Let $y=f(x)$ and $y_0=f(x_0) .$ Then there is a function $\Phi $ continuous at $x_0$ such that $\Phi (x_0)=f’(x_0)\neq 0 $ and $$f(x)=f(x_0) +\Phi (x)(x-x_0). $$ Now since $\Phi $ is continuous at $x_0$ then $\Phi (x)\neq 0$ close enough to $x_0$. So $1/\Phi $ is defined close enough to $x_0$.

Now we have $$f^{-1}(y)=f^{-1} (y_0) +(1/\Phi )(f^{-1}(y))(y-y_0).$$

We know $\Phi f^{-1} $ is continuous at $y_0$ since $f^{-1}$ continuous at $y_0$ and $\Phi $ continuous at $x_0=f^{-1}(y_0)$. So $f^{-1} $ is differentiable at $y_0 $ and $$(f^{-1})’(y_0)=(1/\Phi (f^{-1}(y_0)) = \frac{1}{f’(x_0)}.$$

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  • $\begingroup$ If the inverse is not differentiable then this does not work so does this method evaluation fail? Is it possible that the derivative exists while the method of evaluation fails? $\endgroup$ – Pax Daga Jun 20 '20 at 5:46
  • $\begingroup$ If f inverse is not differentiable then this fails is there an example of such? $\endgroup$ – Pax Daga Jun 20 '20 at 6:01
  • $\begingroup$ If the inverse isn’t differentiable then of course you can’t find it’s derivative. $\endgroup$ – Anon Jun 20 '20 at 12:51
  • $\begingroup$ Ok so if the inverse does not exist then it is undefined such as dx/df where f is the constant function is undefined is that correct? Also how exactly can this help us evaluate derivatives like d(cos(x))/d(sin(x)) $\endgroup$ – Pax Daga Jun 20 '20 at 12:55
  • $\begingroup$ The constant function certainly has no inverse. Say we want to find the derivative of arcsin, then the theorem says if $y_0 = \sin x_0 $ then $\text{arcsin}’(\sin x_0)=1/(\cos(x_0))=1/(\sqrt{1-\sin^2 (x_0)})=1/\sqrt{(1-y_0^2)} $. $\endgroup$ – Anon Jun 20 '20 at 13:14
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I think it is better to understand the result with some more clarity.

Let's first begin with the continuity part:

Theorem 1: Let a function $f:[a, b] \to\mathbb {R} $ be strictly monotone and continuous on $[a, b] $ and let $I=f([a, b]) $ be the range of $f$. Then there exists a function $g:I \to\mathbb {R} $ such that $g$ is continuous on $I$ and $$f(g(x)) =x\, \forall x\in I, g(f(x)) =x\, \forall x\in[a, b] $$

The function $g$ is unique and traditionally denoted by $f^{-1}$ and the important point in the above theorem is that inverse of a continuous function is also continuous. Also observe that if a continuous function is invertible it must also be one-one and continuity combined with one-one nature on an interval forces the function to be strictly monotone. Another point worth remarking is that $I=f([a, b]) $ is also an interval which is either $[f(a), f(b)] $ or $[f(b), f(a)] $ depending upon whether $f$ is increasing or decreasing.

You should be able to prove the above theorem using properties of continuous functions on a closed interval.

Once we are done with the continuity part it is not much difficult to deal with derivatives and we have:

Theorem 2: Let a function $f:[a, b] \to\mathbb {R} $ be strictly monotone and continuous on $[a, b] $. Let $c\in (a, b) $ be such that $f'(c) \neq 0$ and $d=f(c) $. Then the inverse function $f^{-1}$ is differentiable at $d$ with the derivative given by $$(f^{-1})'(d)=\frac{1}{f'(c)}=\frac{1}{f'(f^{-1}(d))}$$

Before coming to the proof of the theorem above it is best to illustrate it via a typical example. So let $f:[-\pi/2,\pi/2]\to\mathbb{R}$ be defined by $f(x) =\sin x$ and the range of $f$ here is $I=[-1,1]$. The derivative $f'(x) =\cos x$ is non-zero in $(-\pi/2,\pi/2)$ and hence the inverse function $f^{-1}$ (usually denoted by $\arcsin$) is differentiable on $(-1,1) $.

To evaluate $(f^{-1})'(x)$ for $x\in (-1,1)$ we need to use a point $y\in(-\pi/2,\pi/2)$ such that $x=f(y) =\sin y$ and we have $$(f^{-1})'(x)=\frac{1}{f'(y)}=\frac{1}{\cos y}=\frac{1}{\sqrt{1-\sin^2y}}=\frac{1}{\sqrt{1-x^2}}$$ The proof of above theorem is based on the definition of derivative. One should note that as per theorem 1 the inverse function $f^{-1}$ is continuous on range of $f$ and in particular at point $d=f(c) $. We have \begin{align} (f^{-1})'(d)&=\lim_{h\to 0}\frac{f^{-1}(d+h)-f^{-1}(d)}{h}\notag\\ &=\lim_{k\to 0}\frac{k}{f(c+k)-f(c)}\notag\\ &=\frac{1}{f'(c)}\notag \end{align} Here we have used $$k=f^{-1}(d+h)-f^{-1}(d)=f^{-1}(d+h)-c$$ so that $$d+h=f(c+k)$$ or $$h=f(c+k) - d=f(c+k) - f(c) $$ and note that by continuity of $f^{-1}$ at $d$ we have $k\neq 0,k\to 0$ as $h\to 0$.

It should be observed that for the result to hold we must ensure that derivative $f'(c) \neq 0$ and $f^{-1}$ is continuous at $d=f(c) $.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Aloizio Macedo Jun 21 '20 at 13:19
  • $\begingroup$ So dx/d(sin(x))=1/ cos(x) for all x and to justify this you can use the fact that arcsin is inverse of sin on [-pi/2,pi/2] for other x you can use different inverse and so it would be for all x is that correct? $\endgroup$ – Pax Daga Jun 28 '20 at 8:02
  • $\begingroup$ @VivaanDaga: yes that's correct. The relation holds whenever $\cos x\neq 0$. Also I don't see why you find this surprising rather than a trivial thing. $\endgroup$ – Paramanand Singh Jun 28 '20 at 10:22
  • $\begingroup$ @ParamanandSingh does the parametric differentiation formula only work if the inverse exists or is there some other proof? $\endgroup$ – Pax Daga Jul 5 '20 at 17:17
  • $\begingroup$ @VivaanDaga: there can be other proofs, but the condition for existence of inverse is a must. $\endgroup$ – Paramanand Singh Jul 6 '20 at 2:00
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The theorem you ask for is true in a completely rigorous generalized framework for derivatives, as described in this post with the desired theorem and the proof. For convenience I shall reproduce the two examples from that post, which show without doubt that conditions often claimed to be needed are in fact not. Obviously, if you choose to work in a limited framework, you may not be able to get the same generalized results, but that merely reveals a limitation of the chosen framework rather than a limitation of the theorem itself. For the precise definitions please refer to the linked post.

First is a function that is differentiable only at $0$ but has a discontinuity in every open interval around $0$. Same goes for its inverse. invertible function whose graph is bounded between two hyperbolas tangent at the origin but has a discontinuity in every open interval around zero

Second is a curve that has a well-defined derivative when it passes through the origin but is not locally bijective there: curve that spirals towards and through the origin with a well-defined tangent at origin and yet performs infinitely many spirals in any open disk around the origin

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