Functions of functions in calculus I have just finished working through a chapter on calculus and have coped OK, answering most of the questions correctly. The chapter dealt with composite functions, along with other topics.
But I just cannot understand this question. I can see we are dealing with functions of functions but they are not set out in a form I can understand. Any help would be appreciated.
In each of the following cases, express $f'$, the derivative of $f$ (with respect to $x$) in terms of $g'$. The number $a$ is constant:
$\begin{align} i) \quad & f(x) = g(x + g(a)) \\
ii) \quad & f(x) = g(a + g(x)) \\
iii) \quad & f(x) = g(x^2) 
\end{align}$
 A: First, I think it will help to point out that these are not functions of functions. A composition of functions isn't a function of functions,$^*$ it's just an ordinary function which you might think about as being "built in two (or more) steps."

OK, now on to the problem itself. Consider a simpler example. Suppose $$f(x)=g(2x).$$Both $f$ and $g$ are ordinary functions; maybe $f(x)=4x^2$ and $g(x)=2x$. We don't know enough about $f$ to compute $f'(x)$, but we do know enough about how $f$ relates to $g$ to compute a relationship between $f'$ and $g'$, via the chain rule:

*

*First, we explicitly write $f$ as a composition of two functions: the outer one is just $g$ itself, and the inner one is $h:x\mapsto 2x$. So $f=g\circ h$.


*That's much more confusing to write than just "$f(x)=g(2x)$," but it makes things clearer for applying the chain rule to get $$f'(x)=g'(h(x))\cdot h'(x)=g'(2x)\cdot 2=2g'(2x).$$ So we've shown $f'(x)=2g'(2x)$; again, this doesn't amount to a computation of $f'$ itself, but it does tell us how to compute $f'$ in terms of $g'$.
Now your cases are a bit more complicated, but they're using the same principle. That first bulletpoint is going to be crucial, so focusing on the first problem for simplicity:

Suppose $f(x)=g(x+g(a))$. What's a function $h$ such that $$f=g\circ h$$ (if you can find this $h$ you'll then be able to just apply the chain rule directly)?


$^*$That said, there is a genuine "function-of-functions" floating around here - namely, composition itself! Composition (of functions $\mathbb{R}\rightarrow\mathbb{R}$, for simplicity) is the operation which takes a pair of functions $h_1,h_2$ and spits out the function $h_1\circ h_2$. So composition is a binary function-of-functions.
A: It is the use of the chain rule. For example
$ii) \quad f(x)=g(x+g(x))$
$$\frac{df}{dx} = \frac{d\big(g(x+g(x))\big)}{d\big(x+g(x)\big)} \frac{d\big(x+g(x)\big)}{dx} = g’(x + g(a))(1 + g’(x))$$
(the big parenthesis are only there to avoid confusion)
