Prove by induction that $\frac{d^m }{dx^m}\Bigl(\frac{d^nf}{dx^n}\Bigl)=\frac{d^{m+n}f}{dx^{m+n}}$

Question

I have to prove by induction that :

$\frac{d^m }{dx^m}\Bigl(\frac{d^nf}{dx^n}\Bigl)=\frac{d^{m+n}f}{dx^{m+n}}$ where $m,n \in \mathbb{N}$

What I tried:

1. For $m=1$ the formula is true :

Starting from this one:

$f^{(n)}(x)=\Bigl(f^{(n-1)}(x)\Bigl)'$ or $\frac{d^{n}f}{dx^{n}}=\frac{d}{dx}\Bigl(\frac{d^{n-1}f}{dx^{n-1}}\Bigl)$ $...(*)$

I can prove that for $m=1$ it holds from the definition that:

$\frac{d}{dx}\Bigl(\frac{d^{n}f}{dx^{n}}\Bigl)=\frac{d^{n+1}f}{dx^{n+1}}$

2.Now I suppose that the formula above holds for some $m>1 \in \mathbb{N}$

$\frac{d^{m}}{dx^{m}}\Bigl(\frac{d^{n}f}{dx^{n}}\Bigl)=\frac{d^{n+m}f}{dx^{n+m}}$ $...(**)$ Inductive Hypothesis

3.Then I prove that it holds for $m+1 \in \mathbb{N}$ too:

$\frac{d^{m+1}}{dx^{m+1}}\Bigl(\frac{d^nf}{dx^{n}}\Bigl)=(*)=\frac{d}{dx}\bigg(\frac{d^{m}}{dx^{m}}\Bigl(\frac{d^{n}f}{dx^{n}}\Bigl)\bigg)=(**)=\frac{d}{dx}\Bigl(\frac{d^{n+m}f}{dx^{n+m}}\Bigl)=(*)=\frac{d^{n+m+1}f}{dx^{n+m+1}}$

My first question is : Is this proof correct?

And my second question is about this formula : $f^{(n)}(x)=\Bigl(f^{(n-1)}(x)\Bigl)'$ or $\frac{d^{n}f}{dx^{n}}=\frac{d}{dx}\Bigl(\frac{d^{n-1}f}{dx^{n-1}}\Bigl)$ , I understand the first one $f^{(n)}(x)=\Bigl(f^{(n-1)}(x)\Bigl)'$ comes by definition of the $n$th derivative, but in the second one it uses differentials, and I know that it is also based on definition but I don't clearly understand how it works with differentials. Anyone who could answer my questions will be highly appreciated .Thank you!

Answer 1

Let us write $Df$ for $\frac{df}{dx}$.

We have to be absolutely clear what we mean by $D^{n}\phi$ where $n$ is a natural number for any function $\phi$. We define this recursively: $$ D^{0}\phi:=\phi; \ \ \ D^{n+1}\phi:=D D^{n}\phi.$$

We are asked to prove for all natural numbers that $D^{m}(D^{n}f)=D^{m+n}f$.

Fix $n$ and proceed by induction.

Root case, $m=0$.

$D^{0} D^{n}f=D^{n}f$ by the definition of $D^{0}$. And $D^{n}f=D^{0+n}f$ by the usual properties of addition of natural numbers.

Inductive step. Assume $D^{m} D^{n}f=D^{m+n}f$ and prove that $D^{m+1}D^{n}f=D^{m+n+1}f$

(a) Now $D^{m+1}D^{n}f=D (D^{m} D^{n} f)$ by the definition of $D^{m+1}(D^{n}f)$.

(b) By the inductive hypothesis, $D^{m} D^{n} f=D^{m+n}f$, so we have $D^{m+1}D^{n}f=D (D^{m+n} f)$.

(c) By the definition of $D^{m+n+1}f$ we have that $D D^{m+n}f =D^{m+n+1}f$.

Putting these together we have that $D^{m+1}D^{n}f=D^{m+n+1}f$.

Answer 2

Your proof of $$\frac{d^m }{dx^m}\Bigl(\frac{d^nf}{dx^n}\Bigl)=\frac{d^{m+n}f}{dx^{m+n}}$$

using mathematical induction is flawless.

On your question about $$f^{(n)}(x)=\Bigl(f^{(n-1)}(x)\Bigl)'$$

in comparison with $$\frac{d^{n}f}{dx^{n}}=\frac{d}{dx}\Bigl(\frac{d^{n-1}f}{dx^{n-1}}\Bigl)$$ They are the same thing.

You may use one or the other as you wish.