Chain rule confusion in defining composite function I am currently a Calc 1 student, and I have learned the Chain rule for differentiation. 
The Chain Rule states that $[f(g(h(x)))]' = f'(g(h)) \times g'(h) \times h'$. I totally understand this. 
However, I do not understand what constitutes as a composite function for the chain rule. For example, with the function $f(x) = (5/(x^2+10))^6$, I can divide this function into three parts, $f(x), g(x)$, and $h(x)$. $f(x) = x^6$, $g(x) = 5/(x+10)$, and $h(x) = x^2$. Therefore, $f(g(h(x))) = (5/(x^2+10))^6$. 
Based on the chain rule, I need to evaluate $f'(g(h)) \times g'(h) \times h'$, which does not get me to the answer. While I understand how to get to the answer, I am confused as to what constitutes as a valid composite function in the chain rule because there are many ways in which one can define a composite function, such as the one above. 
Any help is greatly appreciated!
Here is what I have for the compositions: 
$$f'(x) = 6(5/(x^2+10))^5 \times -5(2x)/(x^2+10)^2 \times 2x$$
Please let me know if I am on the right track!
Thanks,
Daanyal
 A: 
I do not understand what constitutes as a composite function for the chain rule.

An easy way to think of a composite function is to think of a function within a function, i. e., nested functions. For example here is a composite function,
$$f(x) = (g(x))^2$$
You could have written it like this,
$$f(t) = t^2, \quad f(g(x))$$
This composite form $f(g(x))$ is also denoted as $f \circ g (x)$. Chain rule lets us differentiate such functions like this,
$$\frac{\operatorname{d} f}{\operatorname{d} x} = \frac{\operatorname{d} f}{\operatorname{d} g} \times \frac{\operatorname{d} g}{\operatorname{d} x} \Rightarrow \frac{\operatorname{d} f}{\operatorname{d} x}=\frac{\operatorname{d} (g(x))^2}{\operatorname{d} x}=\frac{\operatorname{d} (g(x))^2}{\operatorname{d} g(x)}\frac{\operatorname{d} g(x)}{\operatorname{d} x} = 2g(x)\frac{\operatorname{d} g(x)}{\operatorname{d} x}$$
There squaring was the outer function and $g$ was the inner function.

Now for your function,
$$f(x) = \Bigg(\frac{5}{x^2+10}\Bigg)^6$$
You have correctly identied $f,g$ but let's change $h$ a bit.
$$f(x)=x^6, \quad g(x)=\frac{5}{x}, \quad h(x)=x^2 + 10, \quad f(g(h(x)))=\Bigg(\frac{5}{x^2+10}\Bigg)^6$$
Can you use the chain rule to take the derivative of $f(g(h(x)))$?
\begin{align*}
\frac{\operatorname{d} f(g(h(x)))}{\operatorname{d} x}
& = 
\frac{\operatorname{d} f(g(h(x)))}{\operatorname{d} g(h(x))}
\frac{\operatorname{d} g(h(x))}{\operatorname{d} x}
&& \text{Applied Chain Rule Once} \\
& = 
\frac{\operatorname{d} f(g(h(x)))}{\operatorname{d} g(h(x))}
\frac{\operatorname{d} g(h(x))}{\operatorname{d} h(x)}
\frac{\operatorname{d} h(x)}{\operatorname{d} x}
&& \text{Applied Chain Rule Twice} \\
& = 
\frac{\operatorname{d} f}{\operatorname{d} g}
\frac{\operatorname{d} g}{\operatorname{d} h}
\frac{\operatorname{d} h}{\operatorname{d} x}
&& \text{This looks cleaner}
\end{align*}
Now we can solve each differential individually and take their product.
\begin{align*}
\frac{\operatorname{d} f}{\operatorname{d} g} &= 6g^5 \\
\frac{\operatorname{d} g}{\operatorname{d} h} &= \frac{-5}{x^2} \\
\frac{\operatorname{d} h}{\operatorname{d} x} &= 2x
\end{align*}
Therefore we have,
$$
\frac{\operatorname{d} f}{\operatorname{d} x}
= \frac{\operatorname{d} f}{\operatorname{d} g}
\frac{\operatorname{d} g}{\operatorname{d} h}
\frac{\operatorname{d} h}{\operatorname{d} x}
= 6g(x)^5 \times \frac{-5}{h(x)^2} \times 2x
$$
