1

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

scribe
  • 524
Daanyal Akhtar
  • 21
  • 1
    There are indeed many ways, but they should all yield the same answer. Remember too that when you write $g'(h)$, you mean $g'(h(x)))$. Likewise, $f'(g(h))$ means $f'(g(h(x)))$. It may help if you provide your work so we can see what might have been done wrong. Specifically, what do you have for $f'$, $g'$, and $h'$? – MPW Oct 16 '18 at 17:06
  • Just updated my post. Sorry for the confusion! – Daanyal Akhtar Oct 16 '18 at 17:14
  • 1
    Your mistake is $g'(h)$. It should be $-5 / (x^2 + 10)^2$. Because you shouldn't do $\frac{d}{dx}\frac{5}{x^2 +10}$. Rather you should be doing $\frac{d}{d(x^2)}\frac{5}{x^2 + 10}$ – DWade64 Oct 16 '18 at 17:23
  • Thanks so much DWade64! Sorry, I don't completely understand what you are saying. Why d/dx^2? – Daanyal Akhtar Oct 16 '18 at 17:33
  • You're welcome! As MPW said, $g'(h)$ really means you are taking the derivative, with respect to the output of the $h$ function. The output of the $h$ function is $x^2$, or you can relabel this $u = x^2$ if you like. Therefore $g'(h(x))$ really means $\frac{d}{d(h(x))} = \frac{d}{d(x^2)} = \frac{d}{du}$ and we are doing $\frac{d}{du} \frac{5}{u + 10}$. You don't take the derivative with respect to $x$ until the very end. $h'(x)$ means the derivative of $h$ with respect to $x$. This is the only derivative that you take with respect to $x$. All the others are with respect to outputs – DWade64 Oct 16 '18 at 17:51
  • Thank you so much! I really appreciate it! – Daanyal Akhtar Oct 16 '18 at 17:57

1 Answers1

1

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, $$ \frac{\operatorname{d} f}{\operatorname{d} x} = \frac{\operatorname{d} f}{\operatorname{d} g} \frac{\operatorname{d} g}{\operatorname{d} h} \frac{\operatorname{d} h}{\operatorname{d} x} = 6g(x)^5 \times \frac{-5}{h(x)^2} \times 2x $$

scribe
  • 524