Section 12 Derivative: rules
We now examine how the operation of taking the derivative of a function interacts with our basic function operations: e.g., scalar multiplication, addition, multiplication, and quotients. We will make heavy use of the Leibniz notation \(\frac{d}{dx}\text{,}\) which denotes the operation of taking the derivative with respect to \(x\text{.}\) In this context we will either write \(\frac{d}{dx}(f)\text{,}\) where \(f\) is the name of a function, or else \(\frac{d}{dx}(\text{blah})\text{,}\) where blah denotes some expression in \(x\text{.}\) In the latter case, the expression in question is understood to define a function, with the usual implied domain convention. Thus for example \(\frac{d}{dx}(5)\) denotes the derivative of the constant function \(f(x)=5\text{;}\) and \(\frac{d}{dx}(x)\) denotes the derivative of the identity function \(f(x)=x\text{.}\)
Theorem 12.1. Derivative formulas: constant and power functions.
Example 12.3. Power function formula: radicals.
Let \(f(x)=\sqrt[3]{x^2}\text{.}\) Compute \(f'(x)\text{.}\)
Solution.
The key is to convert the formula of \(f\) from radical to power form:
\begin{equation*}
f(x)=(x^2)^{1/3}=x^{2/3}.
\end{equation*}
We then use the power formula to compute the derivative:
\begin{align*}
f'(x) \amp =\frac{2}{3}x^{-1/3} \amp \text{(power form.)}\\
\amp \frac{2}{3x^{1/3}}\\
\amp = \frac{2}{3\sqrt[3]{x}}\text{.}
\end{align*}
Note that the last two steps, strictly speaking, are unnecessary for this exercise, since no specific instructions were given regarding the form of the final answer. However, it is good to get in the habit of converting fluently between power and radical notation.
Theorem 12.4. Derivative rules.
-
Linear combination.
For all constants \(c,d\in \R\text{,}\) we have
\begin{equation}
\frac{d}{dx}(cf(x)+dg(x))=cf'(x)+dg'(x)\text{.}\tag{12.4}
\end{equation}
-
Product rule.
We have
\begin{equation}
\frac{d}{dx}(f(x)g(x))=f'(x)g(x)+f(x)g'(x)\text{.}\tag{12.5}
\end{equation}
-
Quotient rule.
We have
\begin{equation}
\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f'(x)g(x)-f(x)g'(x)}{\left(g(x)\right)^2}\text{.}\tag{12.6}
\end{equation}
Proof.
With all of our mastery of limit techniques, it turns out that the proofs of these three statements are not so difficult. We provide a proof of (1) and (2).
Proof of (1).
Assume \(f\) and \(g\) are differentiable at \(x\text{.}\) We have
\begin{align*}
(cf(x)+dg(x))' \amp =\lim\limits_{h\to 0}\frac{cf(x+h)+dg(x+h)-(cf(x)+dg(x))}{h}
\amp \text{(def.)}\\
\amp =\lim\limits_{h\to 0}\frac{cf(x+h)-cf(x)}{h}+\frac{dg(x+h)-dg(x)}{h}\\
\amp = c\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}+d\lim\limits_{h\to 0}\frac{g(x+h)-g(x)}{h} \amp \text{(sum, scal. mult.)}\\
\amp = cf'(x)+dg'(x) \amp \text{(def.)}\text{.}
\end{align*}
Proof of (2).
Assume \(f\) and \(g\) are differentiable at \(x\text{.}\) We have
\begin{align*}
(f(x)g(x))' \amp = \lim\limits_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h} \amp \text{(def.)} \\
(f(x)g(x))' \amp = \lim\limits_{h\to 0}\frac{(f(x+h)-f(x))g(x+h)-f(x)(g(x+h)-g(x))}{h} \amp \text{(alg.)} \\
\amp = \lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}\cdot \lim\limits_{h\to 0}g(x+h)
+\lim\limits_{h\to 0}f(x)\cdot\lim\limits_{h\to 0}\frac{g(x+h)-g(x)}{h}
\amp (\text{sum, prod.}) \\
\amp =f'(x)g(x)+f(x)g'(x)\text{,}
\end{align*}
where this last step uses both the definition of the derivative and the fact that
\begin{align*}
\lim\limits_{h\to 0}f(x+h)\amp =f(x) \\
\lim\limits_{h\to 0}g(x)\amp =g(x) \text{,}
\end{align*}
Example 12.7. Derivative rules.
Compute a formula for \(f'\) for the given \(f\text{.}\)
\(f(x)=\anpoly\text{,}\) where \(a_n,a_{n-1},\dots, a_0\) are fixed constants.
\(\displaystyle \displaystyle f(x)=\frac{3x}{4x^2-3x+1}\)
\(\displaystyle \displaystyle f(x)=\frac{(2+\sqrt{x})(x-2)}{\sqrt[3]{x}}\)
Solution.
We compute
\begin{align*}
f'(x) \amp = a_n(x^n)'+a_{n-1}(x^{n-1})'+\cdots +a_1(x)'+(a_0)' \amp \text{(lin. comb.)}\\
\amp = na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots +2a_2x+a_1 \amp \text{.}
\end{align*}
We compute
\begin{align*}
f'(x) \amp = \frac{(3x)'(4x^2-3x+1)-3x(4x^2-3x+1)'}{(4x^2-3x+1)^2} \amp \text{(quot. rule)}\\
\amp = \frac{3(4x^2-3x+1)-3x(8x-3)}{(4x^2-3x+1)^2} \amp \text{(poly rule from above)}\\
\amp = \frac{-12x^2+3}{(4x^2-3x+1)^2}\text{.}
\end{align*}
If we go straight into using derivative rules, we would have to use the product and quotient rules, which would be time consuming. Instead, realizing that powers of \(x\) appear in all terms (above and below), we first do some algebraic preparation:
\begin{align*}
f(x) \amp =\frac{2x+x^{3/2}-2x^{1/2}-4}{x^{1/3}}\\
\amp = 2x^{2/3}+x^{7/6}-2x^{1/6}-4x^{-1/3}\text{.}
\end{align*}
We can now handle the derivative computation just by using the linear combination rule and power function formulas:
\begin{align*}
f'(x) \amp = 2(x^{2/3})'+(x^{7/6})'-2(x^{1/6})'-4(x^{-1/3})' \amp \text{lin. comb.}\\
\amp = \frac{4}{3}x^{-1/3}+\frac{7}{6}x^{1/6}-\frac{1}{3}x^{-5/6}+\frac{4}{3}x^{-4/3} \amp \text{(power form.)}\text{.}
\end{align*}
The first part of
Example 12.7 provides a formula for computing the derivative of a general polynomial. This is deserving of an official result.
Theorem 12.8. Derivative: polynomials.
Given a polynomial \(f(x)=\anpoly\text{,}\) where \(a_k\in \R\) are fixed constants, we have
\begin{equation}
f'(x)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots +2a_2x+a_1\text{.}\tag{12.9}
\end{equation}
Definition 12.9. Higher order derivatives.
Let \(f\) be a function with derivative \(f'=\frac{df}{dx}\text{.}\) Continuing to compute derivatives in succession yields the higher order derivatives of \(f\text{.}\) In particular, the second derivative is defined as \(\frac{d}{dx}(f')\text{,}\) and denoted as \(f''\) or \(\frac{d^2 f}{dx^2}\text{;}\) and the third derivative is defined as \(\frac{d}{dx}(f'')\text{,}\) and is denoted \(f'''\) or \(\frac{d^3 f}{dx^3}\text{.}\)
More generally, for any \(n\geq 1\text{,}\) the \(n\)-th derivative of \(f\) is the result of applying the derivative operation \(n\) times in succession, and is denoted \(f^{(n)}\) or \(\frac{d^n f}{dx^n}\)
Example 12.10. Higher-order derivatives.
Let \(f(x)=x-2x^2\text{.}\) Compute formulas for \(f'\text{,}\) \(f''\text{,}\) and \(f'''\text{.}\)
Solution.
We compute the derivatives in succession:
\begin{align*}
f'(x) \amp = 1-4x \amp \text{(poly. form.)}\\
f''(x) \amp = \frac{d}{dx}(f'(x)) \\
\amp = (1-4x)'\\
\amp = -4 \amp \text{(poly. form.)}\text{.}
\end{align*}
