If $f, g$ are both smooth functions of $t$, why does $f \circ g$ is a smooth function of $t$?

Let $$f,g: \mathbb{R} \to \mathbb{R}$$.

I think it is clear that $$f \circ g$$ has derivatives of all orders since $$\dfrac{d^n}{dt^n} (f\circ g)$$ only depends of $$f, g$$ and it's derivatives. But I'm asking for a formal proof because I can't figure it out one.

Can someone helps with a formal proof?

Thanks.

• Do you want to see by definition.. do you know if f and g are differetiable then thir composition is differentiable – Praphulla Koushik May 18 at 18:30
• Do you know Faa Di Bruno formula ? – Jean Marie May 18 at 20:05

We can prove the following statement by induction on $$n$$: for all $$n\in\mathbb{N}$$, if $$f,g:\mathbb{R}\to\mathbb{R}$$ are $$n$$ times differentiable, then $$f\circ g$$ is $$n$$ times differentiable. The base case $$n=0$$ is trivial.

Now suppose $$n>0$$ and the statement is known for $$n-1$$ and we wish to prove it for $$n$$. Let $$f,g:\mathbb{R}\to\mathbb{R}$$ be $$n$$ times differentiable. By the chain rule, $$f\circ g$$ is differentiable with $$(f\circ g)'=g'\cdot (f'\circ g)$$. Now $$f'$$ and $$g$$ are both $$n-1$$ times differentiable, so by the induction hypothesis, $$f'\circ g$$ is $$n-1$$ times differentiable. Since $$g'$$ is also $$n-1$$ times differentiable, $$(f\circ g)'$$ is $$n-1$$ times differentiable since it is a product of two $$n-1$$ times differentiable functions. Thus $$f\circ g$$ is $$n$$ times differentiable.

(Here I assume you already know that a product of two functions which are $$n$$ times differentiable is $$n$$ times differentiable. If you don't know that, it can be proved by induction on $$n$$ in the same way, using the product rule instead of the chain rule.)

As $$f$$ is cont., we see that $$f(x+t)\rightarrow f(x)$$ as $$t\rightarrow 0$$. So, we have

$$\lim_{t\rightarrow 0}\frac{(g\circ f)(x+t)-(g\circ f)(x)}{t} =\lim_{f(x+t)\rightarrow f(x)}\frac{g(f(x+t))-g(f(x))}{t}$$

But, we don’t know what is this. Arrange this as,

$$\lim_{f(x+t)\rightarrow f(x)}\frac{g(f(x+t))-g(f(x))}{t} =\lim_{f(x+t)\rightarrow f(x)}\frac{g(f(x+t))-g(f(x))}{f(x+t)-f(x)}\cdot\frac{f(x+t)-f(x)}{t}$$

Limit distributes in product (here) and so, we have $$\lim_{f(x+t)\rightarrow f(x)}\frac{g(f(x+t))-g(f(x))}{f(x+t)-f(x)}\lim_{t\rightarrow 0}\frac{f(x+t)-f(x)}{t}=g’(f(x))f’(x)$$

So, the limit exists and so the composition $$g\circ f$$ is differentiable.

Suppose $$g,f$$ are smooth. It is clear that $$g\circ f$$ is differentiable.

$$g’\circ f$$ is differentiable being composition of differentiable functions.

As product of differentiable functions is a differentiable function, $$(g’\circ f)\cdot f’$$ is differentiable, that is, $$(g\circ f)’$$ is differentiable.

Similarly, we can prove every derivative of $$g\circ f$$ is differentiable, that is $$g\circ f$$ is smooth.