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Section 2.5 Chain Rule of Derivatives
An additional property of derivatives is the
chain rule which addresses composition of functions. In addition to being useful for calculating derivatives it enables a variety of applications which will be presented in later sections.
Your goal by the end of this section is to be able to calculate a derivative requiring the chain rule.
Subsection 2.5.1 Illustrating the Chain Rule
Checkpoint 2.5.1 . Illustration of Chain Rule.
Suppose for every shovel full of coal a steam engine produces 1000 psi, and for every 2000 psi the train travels one mile.
What is the rate of psi to coal?
What is the rate of miles to psi?
What is the rate of coal to miles?
Write a function that outputs psi given number of shovels of coal.
Calculate the derivative of this function.
Write a function that outputs miles given psi.
Calculate the derivative of this function.
Write a function that outputs miles given number of shovels of coal using the previous two functions.
Calculate the derivative of this function.
Subsection 2.5.2 Property and Example
Theorem 2.5.2 . Chain Rule.
If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g(x)\) then \(h=f(g(x))\) is differentiable at \(x\) and
\begin{equation*}
h^\prime(x)=f^\prime(g(x))g^\prime(x)
\end{equation*}
Example 2.5.3 .
Calculate the derivative of
\(h(x)=\sqrt{x^2-1}.\)
Solution 1 .
Note how each part of the chain rule is calculated and assembled.
\begin{align*}
h(x) = \amp \sqrt{\color{red}{x^2-1}}\\
f(u) = \amp \sqrt{u}. \\
\color{red}{g(x)} = \amp \color{red}{x^2-1}. \\
f^\prime(u) = \amp \frac{1}{2}u^{-\frac{1}{2}}. \\
\color{red}{g^\prime(x)} = \amp \color{red}{2x}. \\
h^\prime(x) = \amp f^\prime({\color{red}{g(x)}})\color{red}{g^\prime(x)}\\
= \amp \frac{1}{2}({\color{red}{x^2-1}})^{-\frac{1}{2}}\color{red}{2x} \\
= \amp \frac{x}{\sqrt{x^2-1}}.
\end{align*}
Solution 2 .
This is the same solution using a different notation that some people find easy to use. It is designed to help us do one step at a time without losing track of where we are.
\begin{align*}
f(u) = \amp \sqrt{u}. \\
g(x) = \amp x^2-1. \\
h^\prime(x) = \amp \frac{1}{2}(x^2-1)^{-\frac{1}{2}}[x^2-1]^\prime \\
h^\prime(x) = \amp \frac{1}{2}(x^2-1)^{-\frac{1}{2}}2x \\
= \amp \frac{x}{\sqrt{x^2-1}}.
\end{align*}
Example 2.5.4 .
Calculate the derivative of
\(h(x)=\sin(2\pi x)\)
Solution .
\begin{align*}
h(x) = \amp \sin(\color{red}{2\pi x})\\
f(u) = \amp \sin(u). \\
\color{red}{g(x)} = \amp \color{red}{2\pi x}. \\
f^\prime(u) = \amp \cos(u). \\
\color{red}{g^\prime(x)} = \amp \color{red}{2\pi}. \\
h^\prime(x) = \amp f^\prime({\color{red}{g(x)}})\color{red}{g^\prime(x)}\\
= \amp \cos({\color{red}{2\pi x}})\color{red}{2\pi} \\
= \amp 2\pi\cos(2\pi x).
\end{align*}
Example 2.5.5 .
Calculate the derivative of
\(h(y)=\ln(\sin y + \cos y)\)
Solution .
\begin{align*}
h(y) = \amp \ln(\color{red}{\sin y + \cos y})\\
f(u) = \amp \ln(u). \\
\color{red}{g(y)} = \amp \color{red}{\sin y + \cos y}. \\
f^\prime(u) = \amp \frac{1}{u}. \\
\color{red}{g^\prime(y)} = \amp \color{red}{\cos y - \sin y}. \\
h^\prime(y) = \amp f^\prime({\color{red}{g(y)}})\color{red}{g^\prime(y)}\\
= \amp \frac{1}{\color{red}{\sin y + \cos y}}\color{red}{\cos y - \sin y} \\
= \amp \frac{\cos y - \sin y}{\sin y + \cos y}.
\end{align*}
Subsection 2.5.3 Implicit Differentiation
Sometimes we know an equation involving a function before we know what the function is. We can still calculate the derivative of this unknown function. This method (mostly a notation) is vital for
Section 2.6
Example 2.5.6 . First Use of Implicit Differentiation.
Calculate
\(f^\prime(x)\) given
\(f(x)^2+x^2 = 2x+3.\) Notice that this does not provide a rule for the function. Rather it shows us a relationship involving the function.
Solution 1 .
\begin{align*}
f(x)^2+x^2 = \amp 2x+3. \\
2f(x)(f^\prime(x))+2x = \amp 2. \\
2f(x)(f^\prime(x)) = \amp 2-2x. \\
f^\prime(x) = \amp \frac{2-2x}{2f(x)} \\
f^\prime(x) = \amp \frac{1-x}{f(x)}.
\end{align*}
Solution 2 .
The notation we used reminds us that \(f(x)\) is a function and thus we need to treat it differently. However, the notation is unnecessarily bulky. Here we replace it with \(y=f(x)\text{.}\) This notation is common in our application of implicit differentiation in this class and in future classes.
\begin{align*}
y^2+x^2 = \amp 2x+3. \\
2y(y^\prime)+2x = \amp 2. \\
2y(y^\prime) = \amp 2-2x. \\
y^\prime = \amp \frac{2-2x}{2y} \\
y^\prime = \amp \frac{1-x}{y}.
\end{align*}
Checkpoint 2.5.7 .
Note in
Example 2.5.6 we can calculate the function from the first step. Do this then calculate the derivative. Compare this to the result above.
Example 2.5.8 . Another Implicit Differentiation.
Calculate
\(y^\prime\) given
\(y\) is a function of
\(x\) and
\(y^2-5y+6=x.\)
Solution .
\begin{align*}
y^2-5y+6 = \amp x. \\
2y y^\prime -5y^\prime = \amp 1. \\
y^\prime(2y-5) = \amp 1. \\
y^\prime = \amp \frac{1}{2y-5}.
\end{align*}
