Why can chain rule not be applied recursively (reapplied to outer function)?

Question

1. $f(x)=\sqrt{x+1}$
2. $f(x)=(x+1)^{\frac 12}$
3. $\frac {d}{dx}f(x)=\frac {d}{dx}[(x+1)^{\frac 12}]·\frac {d}{dx}(x+1)$
4. $\frac {d}{dx}f(x)\neq[\frac {d}{dx}[(x+1)^{\frac 12}]·\frac {d}{dx} (x+1)]·\frac {d}{dx}(x+1)$

Why can chain rule not be reapplied to $\frac {d}{dx}[(x+1)^{\frac 12}]$ in step three even though it is also a composite function (it looks the same as $f(x)$). Is it just a rule that chain rule can only be applied once? Does Leibniz's notation explain it?

Answer 1

Chain rule is the following:

$$ [f(g(x))]' = f'(g(x))g'(x) $$

So, when we want differentiate things like $\sqrt{1+x}$, we do $f(x) = x^{1/2}$ and $g(x)= 1+x$ . Then, we get

$$f'(x) = \frac{1}{2\sqrt{x}} \implies f'(g(x)) = \frac{1}{2\sqrt{1+x}}$$

$$g'(x) = 1 $$

Replacing in the formula for the chain rule:

$$ [f(g(x))]' = f'(g(x))g'(x) $$ $$ (\sqrt{1+x})' = \frac{1}{2\sqrt{1+x}} $$

Maybe you are confused with the notation.

Answer 2

The chain rule can be applied recursively.

For example, if $a(x) =f(g(h(x))) $, then

$\begin{array}\\ a'(x) &=(f(g(h(x))))'\\ &=(g(h(x))'f'(g(h(x))) \qquad\text{chain rule applied to } f\\ &=h'(x)g'(h(x))f'(g(h(x))) \qquad\text{chain rule applied to } g\\ \end{array} $