Finding an expression for nth derivative

Question

I've been thinking about this problem for a while now, but can't seem to get anywhere with it.

Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ has derivatives of all orders. Prove that the same is true of $F(x) := \exp(f(x))$. Find an expression for $F^{(n)}(x)$ for $n = 0,...,3$

Ok, I've sort of showed the first part by showing that all derivatives of $F$ consist of certain terms which all exist due to the infinitely differentiable $f$ and $\exp(x)$ and by the linearity of differentiation, chain rule and product rule. The problem is, I don't think it's terribly rigorous, but given how the question doesn't want me to give an expression for $F^{(n)}(x)%$ for all $n$, I can't see any other way to make it more rigorous.

The actual problem I'm having though is just finding the expression for $F^{(n)}(x)$ for $n = 0,...,3$

I've got the following so far

$F(x) = \exp(f(x))$

$F'(x) = \exp(f(x))\cdot f'(x)$

$F''(x) = \exp(f(x))[f''(x) + (f'(x))^2]$

$F'''(x) = \exp(f(x))[f'''(x) + 3f'(x)\cdot f''(x) + (f'(x))^3]$

and really can't see how to get an expression from this. Any help in finding an expression or for my earlier doubts would be greatly appreciated!

Thanks

Answer 1

This follows from the fact that the composition of $C^\infty$ functions is $C^\infty$. One way to show this is via Faà di Bruno's formula, which states $${d^n \over dx^n} f(g(x))=\sum \frac{n!}{m_1!\,1!^{m_1}\,m_2!\,2!^{m_2}\,\cdots\,m_n!\,n!^{m_n}}\cdot f^{(m_1+\cdots+m_n)}(g(x))\cdot \prod_{j=1}^n\left(g^{(j)}(x)\right)^{m_j}.$$ where the sum is over all $n$-tuples of nonnegative integers $(m_1,\ldots,m_n)$ satisfying $m_1+2m_2+3m_3+\cdots+nm_n=n$.

A (in my opinion) better way is by induction on $n$: Suppose we've shown that $(f\circ g)^{(n)}(x)$ exists for all $x$ and is a polynomial $P$ in the "variables" $f(g(x)),f'(g(x)),\ldots,f^{(n)}(g(x)),g(x),g'(x),\ldots,g^{(n)}(x)$ (which is clearly true when $n=1$). I'll label these "variables" $u_1,\ldots u_k$, respectively, where $k=2n+2$. Then, assuming it exists, $$(f\circ g)^{(n+1)}(x)= \frac{d}{dx}P(u_1,\ldots,u_k)=\sum_{j=1}^k \frac{\partial P}{\partial u_j}\!(u_1,\ldots,u_k)\,u_j'(x).$$ But this is just a polynomial in the "variables" $f(g(x)),f'(g(x)),\ldots,f^{(n+1)}(g(x)),g(x),g'(x),\ldots,g^{(n+1)}(x)$. Thus, $(f\circ g)^{(n+1)}(x)$ does in fact exist for all $x$ and is equal to a polynomial in the required "variables". Hence, by induction, $f\circ g$ has derivatives of all orders (i.e. is $C^\infty$).

Answer 2

Avi Steiner's answer is perfectly good and gives a general expansion (I would accept that one), but I wanted to add something that might be useful to supplement the intuition a little.

Depending on how abstract you like things, it might help to think of this in terms of operators. The "derivative" operator $D$ operates on a differentiable function like so: $Df = f'$. $D$ is also linear, which helps things a bit. So let's say that you have your function $F = \exp \circ f$. Now suppose that we want to know what $D(FY)$ is, for some differentiable $Y$. (In your case is $Y(x) = 1$, but we won't worry about that for now).

Then $D(FY)=(DF)Y + F(DY)$. But we know that $DF=(\exp\circ f)Df = FDf$. So $D(FY) = F(Df)Y+FDY = F(Df+D)Y$. That last bit is a little weird, but if you accept it, it works well. It just means "add $(Df)Y$ to $DY$."

So now what is $D^2(FY)$? That's $D(D(FY)) = D(F(Df+D)Y)$. But we know what $D(FZ)$ is for differentiable $Z$ -- it's $F(Df+D)Z$. So these just stack up. That is, you get $D^2(FY)=F(Df+D)^2Y$, and inductively, $D^n(FY)=F(Df+D)^nY$.

Now we can apply it to the case where $Y$ is a constant function. So we have

$$\begin{align*} F' &= F(Df + D)1 = FDf \\&= Ff' \\F'' &= F(Df + D)^21 = F((Df)^2 + (Df)D + (D)Df + D^2)1 \\&= F[(f')^2 + f''] \\F''' &= F(Df + D)^31 = F(Df + D)((Df)^2 + D^2f) \\&= F[(Df)^3 + (Df)D^2f+D((Df)^2) + D^3f] = F[(Df)^3 + 3(Df)D^2f + D^3f] \\&= F[f'^3 + 3f'f'' + f'''] \\\dots \end{align*}$$

And you can keep going like this. One important thing to remember though, is that the operators aren't commutative, that is, $DfD \neq (D)Df$, so you can't just use the binomial formula. But it might give you a good idea about how to see the structure.

Answer 3

Another more concise version of Faà di Bruno's formula:

$$\partial^n(f\circ g)=\sum_{P \text{ an ordered partition of }n}(\partial^{|P|}f)\circ g\cdot\prod_{i\in P}\partial^ig$$ where $|P|$ is the number of terms in the partion $P$ and $\partial^n$ is a convenient shorthand for $\frac{1}{n!}\frac{d^n}{dx^n}$.

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Here is an example of this in action with $n=4$.

There is only one ordered partition $P$ of length 1, namely $\{4\}$. So we have one term $(\partial^1f)\circ g\cdot\partial^4 g$.

There are three partitions of length 2: $\{2,2\}$ and two orderings of $\{3,1\}$. These give us terms $(\partial^2f)\circ g\cdot(\partial^2 g)^2$ and $2(\partial^2f)\circ g\cdot(\partial^3 g)(\partial^1 g)$.

There are three partitions of length 3: three orderings of $\{2,1,1\}$. These give us the term $3(\partial^3f)\circ g\cdot\partial^2 g\cdot(\partial^1 g)^2$.

And there is only one partition of length 4: $\{1,1,1,1\}$. This gives us the term $(\partial^4f)\circ g\cdot(\partial^1 g)^4$

So $\partial^4(f\circ g)=(\partial^1f)\circ g\cdot\partial^4 g+(\partial^2f)\circ g\cdot(\partial^2 g)^2+2(\partial^2f)\circ g\cdot(\partial^3 g)(\partial^1 g)+3(\partial^3f)\circ g\cdot\partial^2 g\cdot(\partial^1 g)^2+(\partial^4f)\circ g\cdot(\partial^1 g)^4$

If we like, we can convert to the standard $\frac{d}{dx}$ by introducing the right factorials. In the line that follows I have then multiplied across by $4!$ and simplified.

$\frac{d^4}{dx^4}(f\circ g)=(\frac{d}{dx}f)\circ g\cdot\frac{d^4}{dx^4} g+3(\frac{d^2}{dx^2}f)\circ g\cdot(\frac{d^2}{dx^2} g)^2+4(\frac{d^2}{dx^2}f)\circ g\cdot(\frac{d^3}{dx^3} g)(\frac{d}{dx} g)+6(\frac{d^3}{dx^3}f)\circ g\cdot\frac{d^2}{dx^2} g\cdot(\frac{d}{dx} g)^2+(\frac{d^4}{dx^4}f)\circ g\cdot(\frac{d}{dx} g)^4$