Proof of dy=f’(x)dx

Question

I’ve been wondering about the usage of $dy=f’(x)dx$ in my textbook.

There’s not a single justification of how it is proved and it just states that it is true.

Since $dy/dx$ can’t be assumed as a fraction, I’m guessing there’s more to it than just multiplying by $dx$ on both sides.

Are there any proofs to this equation?

Also with some research, I found this “proof”. Can it be done this way?

$dy=f’(x)dx$ “proof”

(Please keep this in high school level)

Answer 1

They key to the question is: What do we mean by $dx$ and $dy$, specifically?

In the end, the definition is the thing, and a real definition is more complicated that one would like.

More generally, if $y=f(x)$ with $f$ differentiable, and $g$ is any "nice" function then:

$$\int_{f(a)}^{f(b)} g(y)\,dy = \int_{a}^{b} g(f(x))f'(x)\,dx\tag{1}$$

when the right side is defined.

This is basically saying we can do substitution to compute $\int_{a}^{b} g(f(x))f'(x)\,dx.$

When $g(y)=1$ for all $y$ this means:

$$\int_{f(a)}^{f(b)} dy = \int_{a}^{b} f'(x)\,dx$$

Unfortunately, this doesn't quite look like $dy=f'(x)\,dx$ because the ranges are different.

In intro calculus, you can probably prove (1). When you define the more general Riemann-Stieltjes intregral, there is a notion of $df$ for integrals. and you'll get:

$$\int_a^b h(x)\,df = \int_a^b h(x)f'(x)\,dx$$ when $f$ is continuously differentiable.

There are more advanced notions, like differential forms, but those are specifically defined just to have this and other properties that you'd expect if you wanted to treat $dx$ and $dy$ as algebraic "things."

In the differentiable form view, we have, for any interval $[a,b]$, a curve in 2D space, $C_{a,b}$, the set of points $(x,f(x))$ for $x\in[a,b].$ Then $dy$ is a form on that curve, and $f'(x)dx$ is another form, and both forms evaluate to the same thing on this curve.

This makes clear that this has to do with a curve in 2-dimensional space, and that is why the ranges in (1) seem to shift - we are actually looking at the boundary points on the curves, $(a,f(a))$ and $(b,f(b)).$

But that view is way above intro calculus.

Answer 2

The proof you linked is no better than the proof using fraction notation.

There's two points of view about these matters.

One is the point of view taught in a first calculus course: calculation of derivatives and integrals using these symbolic methods leads to correct answers (which is not hard to prove even in a first calculus course), so learn how to do it.

Another point of view is what is taught in a more advanced course, namely differential geometry: the symbols $dx$ and $dy$ and $f'(x) \, dx$ indeed do have independent mathematical meaning, they are examples of differential forms, and if $y=f(x)$ then $dy = f'(x) dx$ can be proved as a consequence of theorems about differential forms.

Answer 3

Well the derivative is given by: $$\lim_{dx \to 0} \frac{f(x+dx)-f(x)}{dx}=\lim_{dx\to 0} \frac{dy}{dx}$$ By definition the derivative is the rate of change of y with regard to x. That's why RHS stands. As you realise $\frac{dy}{dx}$ is not just a notation but it's mathematically how derivative is been defined. Since $f'(x)=\frac{dy}{dx}$ with $dx\to 0$, the equation $dy=f'(x)dx$ holds.