A calculus of the absurd
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10.5 The chain rule
The chain rule is used to find the derivatives of "functions of a function". Mathematically, these are written as
\[\frac {d}{dx}[y(u(x))]\]
and it’s possible to find this just by known the derivatives of \(y(u)\) and \(u(x)\).
The proof is a little involved, so for the moment you can find it at http://kruel.co/math/chainrule.pdf.
The result we’re after is that
\(\seteqnumber{0}{10.}{32}\)
\begin{equation}
\frac {dy}{dx} = \frac {dy}{du} \frac {du}{dx}
\end{equation}
What is the first thing to do when approaching a question (well after having considered that the chain rule might be relevant)? We must find the nested functions. I find it helpful to think about the parts I find hard to differentiate. For example, in our example, I know how to differentiate
\(\frac {1}{\text {some variable}}\), and also \(\sqrt {x} + 2\) but not
\(\seteqnumber{0}{10.}{34}\)
\begin{equation}
f(x) = \frac {1}{\sqrt {x} + 2}.
\end{equation}
Therefore, it is not unreasonable to create a variable \(u := \sqrt {x} + 2\) and try to use the chain rule. We can write
\(\seteqnumber{0}{10.}{35}\)
\begin{equation}
f(x) = \frac {1}{u}
\end{equation}
This can be a little confusing, because \(x\) no longer seems to appear in the function, but in reality \(u\) is implicitly a function of \(x\) (i.e. we know that \(u\) depends on \(x\)). Applying the chain rule we know that
\(\seteqnumber{0}{10.}{36}\)
\begin{align}
\frac {df}{dx} = \frac {d}{du} \left (\frac {1}{u}\right ) \times \frac {d}{dx} (\sqrt {x} + 2)
\end{align}
We now know how to differentiate all the individual pieces. First we know that
\(\seteqnumber{0}{10.}{37}\)
\begin{align}
\frac {d}{du} \left (\frac {1}{u}\right ) &= \frac {d}{du} \left (u^{-1}\right ) \\ &= -u^{-2} \\ &= -\frac {1}{u^2} \\ &= -\frac {1}{(\sqrt {x} + 2)^2}.
\end{align}
Then we can also differentiate \(\sqrt {x} + 2\) with respect to \(x\), which is just
\(\seteqnumber{0}{10.}{41}\)
\begin{equation}
\frac {1}{2} \frac {1}{\sqrt {x}}
\end{equation}
Therefore we get that overall
\(\seteqnumber{0}{10.}{42}\)
\begin{equation}
\frac {df}{dx} = -\frac {1}{2} \times \frac {1}{\sqrt {x}} \times \frac {1}{(\sqrt {x} + 2)^2}
\end{equation}