# Proving $\frac{dy'}{dy} = \frac{y''}{y'}$ in general without abuse of notation

My understanding of derivatives is that:

$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

Where the limit is defined with the usual $$\epsilon$$-$$\delta$$ style statement with first order logic.

And so $$\frac{df(x)}{dx} = f'(x)$$, as per usual.

This doesn't work so well when people start talking about $$\frac{df'(x)}{df(x)}$$.

In the special case where $$y = f(x)$$ is invertible we can rephrase this with the chain rule :

Let $$g(y) = f^{-1}(y) = x$$, then $$\frac{df'(x)}{df(x)}$$ is:

\begin{align*} \frac{d}{dy}\left( f'(g(y)) \right) &= f''(g(y)) \cdot \frac{d}{dy}( g(y)) \\ &= f''(x) \cdot g'(y) \\ &= f''(x) \cdot g'(f(x)) \\ &= f''(x) \cdot \left( \frac{d}{dx}\left(g(f(x))\right) \cdot \frac{1}{f'(x)} \right) \text{ by chain rule } g'(h(x)) = \frac{d}{dx}( g(h(x)) ) \cdot \frac{1}{h'(x)} \\ &= f''(x) \cdot \left( \frac{d}{dx}\left( x\right) \cdot \frac{1}{f'(x)} \right) \text{ since g(f(x)) = f^{-1}(f(x))}\\ &= \frac{f''(x)}{f'(x)} \end{align*}

It is asserted by many on MSE that: $$\frac{df'(x)}{df(x)} = \frac{f''(x)}{f'(x)}$$ in general.

However I can't seem to make sense of this in terms of the usual $$\epsilon-\delta$$ definitions.

This leads me to think that there are multiple notions of derivatives:

• The ordinary derivative defined with $$\epsilon-\delta$$.
• The notion of a differential, which builds on top of the ordinary derivative.
• Here $$df_x(t) = f'(x) \cdot t$$, where $$d$$ operates on a function. Thus being capable of proving the above derivative trivially: $$\frac{df'_x(t)}{df_x(t)} = \frac{f''(x)}{f'(x)}$$

• I suspect this is what many of the answers are using, and this what people mean when they say "using Leibniz notation"?

My question is the following:

• Is it possible to prove the general case without using the notion of differentials?
• Is it wrong in thinking there are multiple notions of derivatives and that "differentiating with respect to a function" is not the same thing as the ordinary $$\epsilon-\delta$$-based derivative?

Edit: Here are some MSE answers which claim this is true in general:

• Not an answer at all but two comments as you flag this "notation". You should not write $(f(x))'$, the correct notation is $f'(x)$. You differentiate functions, not values of functions. And secondly you don't differentiate with respect to a function $f(x)$, this is an abuse of notation, and one that often leads to confusion. May 12 at 6:53
• What you want to prove "in general" makes no sense in general. The equality is proved at points where you can write the derivative as a function of the function: for example, at points where the function has nonzero derivative (and is therefore locally bijective) May 12 at 8:11
• If f is a function like $x\mapsto x^2$ at the origin, then it is simply not true that the value of the derivative is determined by the value of the function. May 12 at 8:12
• A suggestion: if you are trying to study calculus, I'd suggest you stop worrying about whether the definition of the derivative can be formalized in first order logic and the rest of that... May 12 at 8:16
• @SomeDinosaur I think that most of the posts you reference are to be understood "in the generous spirit of applied mathematics". On the whole they don't distinguish between functions and expressions, and use meaningless conversational fillers like "independent variable". Valuable no doubt in their place, but not place I want to go to. May 12 at 13:35

In the first place, what does $$\frac{\mathrm{d} f' (x)}{\mathrm{d} f (x)}$$ mean? Clearly, it suffices to define what $$\frac{\mathrm{d} g (x)}{\mathrm{d} f (x)}$$ means: once we know that, we can simply substitute $$f'$$ for $$g$$. We should also define it in such a way that when $$f$$ is the identity function – i.e. $$f (x) = x$$ – then $$\frac{\mathrm{d} g (x)}{\mathrm{d} f (x)}$$ has the same meaning as $$\frac{\mathrm{d} g (x)}{\mathrm{d} x}$$. There is an obvious choice: $$\frac{\mathrm{d} g (x)}{\mathrm{d} f (x)} = \lim_{h \to 0} \frac{g (x + h) - g (x)}{f (x + h) - f (x)}$$ Since $$f (x + h) - f (x)$$ could be $$0$$ for $$h \ne 0$$ we should be a little bit more careful, so let us say that the value of $$\frac{\mathrm{d} g (x)}{\mathrm{d} f (x)}$$ at $$x = x_0$$ is $$M$$ if, for all $$\epsilon > 0$$, there exists $$\delta > 0$$ such that for all $$h$$ such that $$0 < \left| h \right| < \delta$$, $$\left| g (x_0 + h) - g (x_0) - M \cdot (f (x_0 + h) - f (x_0)) \right| < \epsilon \cdot \left| f (x_0 + h) - f (x_0) \right|$$

(This definition generalises straightforwardly to the vector-valued multivariable case, provided we understand $$M$$ needs to be a matrix of the appropriate dimensions.) Since both the left and right hand side are non-negative, if $$\frac{\mathrm{d} g (x)}{\mathrm{d} f (x)}$$ has a value at $$x = x_0$$, then there exists $$\delta > 0$$ such that for all $$h$$ such that $$0 < \left| h \right| < \delta$$, $$\left| f (x_0 + h) - f (x_0) \right| > 0$$, i.e. $$f$$ is not constant on any neighbourhood of $$x_0$$.

Now, with all that preamble out of the way, let me state:

Theorem. If $$f (x)$$ and $$g (x)$$ are differentiable at $$x = x_0$$, with $$f' (x_0)$$ and $$g' (x_0)$$ as the values of $$\frac{\mathrm{d} f (x)}{\mathrm{d} x}$$ and $$\frac{\mathrm{d} g (x)}{\mathrm{d} x}$$ at $$x = x_0$$ respectively, and $$f' (x_0) \ne 0$$, then $$\frac{\mathrm{d} g (x)}{\mathrm{d} f (x)}$$ has value $$\frac{g' (x_0)}{f' (x_0)}$$ at $$x = x_0$$.

Proof. Let $$0 < \epsilon < 1$$. By hypothesis, there exists $$\delta_1 > 0$$ such that for all $$h$$ such that $$0 < \left| h \right| < \delta_1$$, $$\left| f (x_0 + h) - f (x_0) - f' (x_0) \cdot h \right| < \frac{1}{3} \epsilon \cdot \left| h \right| \cdot \frac{\min \left\{ \left| f' (x_0) \right|, \left| f' (x_0) \right|^2 \right\}}{\max \left\{ 1, \left| g' (x_0) \right| \right\}}$$ (Replace $$\epsilon$$ with $$\frac{1}{3} \epsilon \cdot \frac{\min \left\{ \left| f' (x_0) \right|, \left| f' (x_0) \right|^2 \right\}}{\max \left\{ 1, \left| g' (x_0) \right| \right\}}$$ in the definition.) We then have: $$\left| f (x_0 + h) - f (x_0) - f' (x_0) \cdot h \right| < \frac{1}{3} \left| f' (x_0) \cdot h \right|$$ $$\left| \frac{g' (x_0)}{f' (x_0)} \right| \cdot \left| f (x_0 + h) - f (x_0) - f' (x_0) \cdot h \right| < \frac{1}{3} \epsilon \cdot \left| f' (x_0) \cdot h \right|$$

Similarly, by hypothesis, there exists $$\delta_2 > 0$$ such that for all $$h$$ such that $$0 < \left| h \right| < \delta_2$$, $$\left| g (x_0 + h) - g (x_0) - g' (x_0) \cdot h \right| < \frac{1}{3} \epsilon \cdot \left| f' (x_0) \cdot h \right|$$ (Replace $$\epsilon$$ with $$\frac{1}{3} \epsilon \cdot \left| f' (x_0) \right|$$ in the definition.)

Let $$\delta = \min \{ \delta_1, \delta_2, 1 \}$$. Then, for all $$h$$ such that $$0 < \left| h \right| < \delta$$: $$\begin{multline} \left| g (x_0 + h) - g (x_0) - \frac{g' (x_0)}{f' (x_0)} \cdot ( f (x_0 + h) - f (x_0) ) \right| \\ \le \left| g (x_0 + h) - g (x_0) - \frac{g' (x_0)}{f' (x_0)} \cdot f' (x_0) \cdot h \right| + \left| \frac{g' (x_0)}{f' (x_0)} \cdot ( f (x_0 + h) - f (x_0) - f' (x_0) \cdot h ) \right| \end{multline}$$ The first term is $$< \frac{1}{3} \epsilon \cdot \left| f' (x_0) \cdot h \right|$$. The second term is also $$< \frac{1}{3} \epsilon \cdot \left| f' (x_0) \cdot h \right|$$. Thus the LHS is $$< \frac{2}{3} \epsilon \cdot \left| f' (x_0) \cdot h \right|$$. But, $$\begin{multline} \left| f' (x_0) \cdot h \right| \le \left| f (x_0 + h) - f (x_0) \right| + \left| f (x_0 + h) - f (x_0) - f' (x_0) \cdot h \right| \\ < \left| f (x_0 + h) - f (x_0) \right| + \frac{1}{3} \left| f' (x_0) \cdot h \right| \end{multline}$$ so $$\left| f' (x_0) \cdot h \right| < \frac{3}{2} \left| f (x_0 + h) - f (x_0) \right|$$. Therefore, $$\left| g (x_0 + h) - g (x_0) - \frac{g' (x_0)}{f' (x_0)} \cdot ( f (x_0 + h) - f (x_0) ) \right| < \epsilon \cdot \left| f (x_0 + h) - f (x_0) \right|$$ as required.　◼

I also beg to differ with those who say that things like "independent variable" are meaningless conversational filler. Pure mathematicians – some of us anyway – know how to make this rigorous. The trick is to recognise the concept of context and make it a concrete thing. In probability theory, this is the purpose of the sample space. We can do the same for basic analysis... but this kind of formalisation is usually not helpful for early students, so we do not teach it.

• That's pretty neat, after following through the proof, we've basically "formalised" what would've been "abuse of notation". This answer my questions, though this does mean $\frac{df(x)}{dg(x)}$ is not defined by "default" for derivatives I presume? Also it seems to have entirely removed the invertible condition which is interesting. In fact we haven't assumed anything other than differentiable and $f'(x_0) ≠ 0$, which seems pretty powerful? May 14 at 16:03
• Indeed, $\frac{\mathrm{d} g (x)}{\mathrm{d} f (x)}$ is not normally defined. This is not the only possible definition but it is perhaps the simplest. May 14 at 22:06

Let us operate with the usual $$\epsilon-\delta$$ definition of derivative. It is standard within this framework to prove (without any handwaving) the usual "rules of differentiation": Linearity, Product Rule, Chain Rule.

Suppose then we have open intervals $$X,Y$$ and twice-differentiable functions $$f:X\to Y$$, $$g:Y\to X$$ satisfying $$f\circ g=I_Y$$ and $$g\circ f=I_X$$ where $$I_X, I_Y$$ are the identity maps on $$X$$, $$Y$$. Suppose that $$f'\not=0$$ on $$X$$ and $$g'\not=0$$ on $$Y$$.

We then have by applying the Chain Rule to $$f\circ g=I_Y$$ that $$(f'\circ g) g'=1, \tag{*}$$ where $$1$$ denotes the constant function whose value is always $$1$$.

From $$(*)$$ we get using the Product Rule that $$(f'\circ g)'g'+(f'\circ g)g''=0$$ which we can rewrite using $$(*)$$ as $$(f'\circ g)'=-\frac{g''}{(g')^2}.\tag{1}$$

But we also have from $$(*)$$ using the Chain Rule and the Product Rule that $$(f''\circ g) (g')^2+(f'\circ g)g''=0$$ which we can rewrite as $$\frac{g''}{(g')^2}=-\frac{(f''\circ g)}{(f'\circ g)}.\tag{2}$$

From (1) and (2) we then have $$(f'\circ g)'=\frac{(f''\circ g)}{(f'\circ g)}. \tag{3}$$

Let us now re-write this in old-fashioned language, evaluating each side at the point $$y\in Y$$, and writing $$x=g(y)$$ (so that $$y=f(x)$$).

The right hand side is clearly just $$\frac{f''(x)}{f'(x)}$$.

The left hand side can be re-written, if we allow ourselves to abuse notation, as follows. The outermost derivative is with respect to $$y=f(x)$$: so write it as $$\frac{d}{df}$$. The innermost derivative is with respect to $$x$$, so let's continue to write it as $$f'$$. Putting this together we have $$\frac{d f'(x)}{df}=\frac{f''(x)}{f'(x)}.$$

If $$(u,v)$$ are functions of $$x$$, we can even bring in another new independent variable $$t$$

$$\frac{du}{dv} = \frac{du/dt}{dv/dt}$$

or continue, permitted by Leibnitz validity with the old independent variable $$x$$ itself by letting

$$u=f'(x), v= f(x)$$

$$\frac{df'(x)}{df(x)} = \frac{\dfrac{df'(x)}{dt}}{\dfrac{df(x)}{dt}}== \frac{\dfrac{df'(x)}{dx}}{\dfrac{df(x)}{dx}}=\frac{f''(x)}{f'(x)}$$

You can even introduce a third independent variable if you wish to.

Btw, using the L'Hôpital Rule or the Quotient Rule if LHS is constant then the RHS would also be the same constant.

• I've never seen this technique of "bringing in another new independent variable"? I'm not really sure what it means and why it should work? Are we multiplying by the ratio of differentials: $\frac{dt}{dt}$ which I presume equals 1? Is this abusive notation? May 12 at 11:56
• Yes, this makes no difference to the old derivative while accommodating any new arbitrary independent variable $t$ May 12 at 12:01
• It is a standard valid non abusive procedure/ trick/artifice to derive relationships among/between differentials in differential geometry. This trick often helps to build new ODEs by computing ratios of segments... as if they are algebraic segment lengths, agreeing with such earliest considerations by Leibnitz, Newton. May 12 at 12:13
• The "derivative" shown here seems to be different from the very basic $\epsilon-\delta$ limit based derivative, with only a single variable. Is the trick here using the concept of differentials? Could this same trick be done without Leibniz notation or differential operators? May 12 at 12:17
• Bringing in a new independent variable is just meaningless and not valid. Nor is this a valid move in differential geometry. May 13 at 4:30