Theorem 3.3.1.
The derivative of the linear function \(f(x)=mx+c\) is
\begin{equation*}
f'(x) = (mx+c)' = \frac{d}{dx}(mx+c) = m.
\end{equation*}
Section 3.3 Differentiation Rules
References.
- Calculus, Early Transcendentals by Stewart, Sections 3.1 and 3.2.
Derivatives of Linear Functions.
Since the slope of a straight line \(y=mx+c\) is the constant \(m\text{,}\) it is easy to check that the derivative of \(f(x)=mx+c\) is \(m\text{,}\) for any constants \(m\) and \(c\text{.}\) It is often convenient to write calculations directly with formulas, without naming the functions, so to illustrate several notations:
The two most basic special cases are when the function is a constant \(c\) or just \(x\text{:}\)
\begin{equation*}
(c)' = \frac{d}{dx}(c) = 0, \quad (x)' = \frac{d}{dx}(x) = 1.
\end{equation*}
Derivatives of Power Functions.
We have already computed the derivatives of a few powers functions like \((x^2)' = 2x\) and \((x^3)' = 3x^2\text{,}\) and these fit a more general pattern:
Theorem 3.3.2. The Power Rule (for Derivatives).
For any non-negative integer \(n\text{,}\)
\begin{equation*}
(x^n)' = \frac{d}{dx}(x^n) = n \cdot x^{n-1}.
\end{equation*}
This also agrees with the results seen above for \(f(x)=x^1 = x\) and \(f(x) = x^0=1\text{.}\)
Note that there is also a “Power Rule for Limits”: from now on, when we simply say “power rule”, we mean this one for derivatives.
To see the pattern that helps us to get the general rule, let us look at \(n=4\text{:}\)
Example 3.3.3.
Calculate the derivative of \(f(x)=x^4\text{.}\)
Solution.Use the first formula for the derivative \(f'(a)\text{:}\)
\begin{equation*}
f'(a) = \lim_{x \to a}\frac{f(x) - f(a)}{x - a}. \lim_{x \to a}\frac{x^4 - a^4}{x - a}.
\end{equation*}
The numerator vanishes for \(x = a\text{,}\) so it has a factor \(x - a\text{,}\) and in fact the factorization is \(x^4 - a^4 = (x-a)(x^3+x^2 \cdot a + x \cdot a^2 + a^3)\text{.}\) (Check by expanding!)
This gives
\begin{align*}
f'(a)
\amp = \lim_{x \to a}\frac{(x-a)(x^3+x^2 \cdot a + x \cdot a^2 + a^3)}{x-a}\\
\amp = \lim_{x \to a} (x^3+x^2 \cdot a + x \cdot a^2 + a^3)\\
\amp = 4a^3.
\end{align*}
That is, \(f'(x) = (x^4)' = 4x^3\text{,}\) as the Power Rule above says.
Let us try the power rule, using various different notations for derivatives:
Example 3.3.4.
For \(f(x) = x^6\text{,}\) find \(f'(x)\text{.}\)
Answer.\(f'(x) = 6x^5\)
Example 3.3.5.
For \(y = x^{1000}\text{,}\) find \(y'\text{.}\)
Answer.\(y' = 1000x^{999}\)
Example 3.3.6.
For \(y = t^{4}\text{,}\) find \(\frac{dy}{dt}\text{.}\)
Answer.\(\frac{dy}{dt} = 4t^{3}\)
Example 3.3.7.
Find \(\frac{d}{dr}(r^3)\text{.}\)
Answer.\(\frac{d}{dr}(r^3) = 3r^{2}\)
Proof of the Power Rule.
The key step is the factorization
\begin{equation*}
x^n-a^n = (x-a)(x^{n-1} + x^{n-2} a + x^{n-3} a^2 + \cdots + xa^{n-2}+ a^{n-1})
\end{equation*}
This can be checked by expanding the right hand side, distributing the left hand factor:
\begin{align*}
\amp (x-a)(x^{n-1} + x^{n-2}a + x^{n-3} a^2 + \cdots + xa^{n-2} + a^{n-1})\\
\amp = x(x^{n-1} + x^{n-2} a + x^{n-3} a^2 + \cdots + xa^{n-2} + a^{n-1})
- a(x^{n-1} + x^{n-2} a + x^{n-3} a^2 + \cdots + xa^{n-2} + a^{n-1})\\
\amp = x^n + x^{n-1} a + x^{n-2}a^2 + \cdots + x^{2}a^{n-2} + xa^{n-1}
- x^{n-1}a - x^{n-2} a^2 - x^{n-3} a^3 - \cdots - xa^{n-1} - a^n\\
\amp = x^n - a^n
\end{align*}
because all the terms in between pair off and cancel out.
Much as with \(x^{4}\text{,}\) the definition of the derivative gives the derivative of \(f(x)=x^n\) at \(x=a\) as
\begin{align*}
f'(a) \amp = \lim_{x \to a}\frac{x^n-a^n}{x-a}\\
\amp = \lim_{x \to a}\frac{(x-a)(x^{n-1} + x^{n-2}\cdot a + x^{n-3}\cdot a^2 + \cdots + a^{n-1})}{x-a}\\
\amp = \lim_{x \to a}(x^{n-1} + x^{n-2} \cdot a + x^{n-3} \cdot a^2 + \cdots + a^{n-1})\\
\amp = a^{n-1} + a^{n-2} \cdot a + a^{n-3} \cdot a^2 + \cdots + a^{n-1}\\
\amp = a^{n-1} + a^{n-1} + a^{n-1} + \cdots + a^{n-1} \text{ (} n \text{ copies)}\\
\amp = n a^{n-1},
\end{align*}
so \(f'(x) = (x^n)' = nx^{n-1}\text{,}\) as claimed.
Constant Multiples, Sums and Differences.
As with limits, we can build up polynomials from these power functions using constant multiples, sums and differences. The derivatives of these three basic combinations are as simple as with limits:
Theorem 3.3.8. The Constant Multiple Rule.
If a differentiable function \(f\) is multiplied by a constant \(c\text{,}\) this product is also differentiable, with derivative:
\begin{equation*}
\frac{d}{dx}[cf(x)] = c\frac{d}{dx}[f(x)], \text{ or } (cf)'(x) = c f'(x).
\end{equation*}
Theorem 3.3.9. The Sum Rule.
The sum of two differentiable functions \(f\) and \(g\) is differentiable, with the sum's derivative the sum of their derivatives:
\begin{equation*}
\frac{d}{dx}[f(x)+g(x)] = \frac{d}{dx}f(x) + \frac{d}{dx}g(x) , \text{ or } (f+g)'(x) = f'(x) + g'(x).
\end{equation*}
Theorem 3.3.10. The Difference Rule.
The difference of two differentiable functions \(f\) and \(g\) is differentiable, with its derivative the difference of their derivatives:
\begin{equation*}
\frac{d}{dx}[f(x)-g(x)] = \frac{d}{dx}f(x) - \frac{d}{dx}g(x) , \text{ or } (f-g)'(x) = f'(x) - g'(x).
\end{equation*}
Checkpoint 3.3.11.
Compute the derivative of \(y=x^8 + 12x^5 - 4x^2 + 10x^3 - 6x + 5\text{.}\) Note that this derivative is also a polynomial.
The same approach works for differentiating any polynomial:
All polynomials are differentiable, and their derivatives are polynomials, so the second and higher derivatives also exist.
Checkpoint 3.3.12.
Find the points on the curve \(y=x^4 - 6x^2 + 4\) where the tangent is horizontal.
Derivatives of Other Power Functions.
Example 3 in Section 2.8 shows that \(\sqrt{x}\) has derivative \({1}/(2 \sqrt{x})\text{.}\) That is, \(\frac{d}{dx}{x^{1/2}} = (1/2)x^{1/2-1}\text{.}\) This fits the power rule, but for power \(1/2\text{,}\) not a positive integer. In fact, this works for all real powers:
Theorem 3.3.13. The Power Rule, Generalized Version.
For any real number \(a\text{,}\)
\begin{equation}
\frac{d}{dx}x^a = a x^{a-1}, \text{ or } (x^a)' = ax^{a-1}.\tag{3.3.1}
\end{equation}
Example 3.3.14.
If \(f(x) = {1}/{x^2}\text{,}\) find \(f'(x)\text{.}\)
Answer.\(f'(x) = {-2}/{x^3}\)
Example 3.3.15.
If \(y = \sqrt[3]{x^2}\text{,}\) find \(y'\text{.}\)
Answer.\(y' = {3}/(2 \sqrt[3]{x})\)
Checkpoint 3.3.16.
Differentiate \(3/x\) .
To simplify differentiation of more functions, we would like to be able to deal with products, quotients, compositions and inverses, much as we did with limits and continuity.
Warning: none of these derivatives are given by rules quite as simple as for limits. Only sums, differences and constant multiples work that simply.
Example 3.3.17. A: a cautionary one.
Compare
- the derivative of the product \(x \cdot x^2=x^3 \)
- the product of the derivatives of \(x \) and \(x^2 \text{.}\)
The Power Rules tells us that the derivative of this product is \(3x^2 \text{;}\) the product of the derivatives of \(x \) and \(x^2 \) is \(1 \cdot 2x = 2x \text{:}\) not the same!
The Derivative of a Product of Functions.
The Leibniz notation is nice here. Let \(u=f(x) \text{,}\) \(v=g(x) \text{,}\) and compute the derivative of the product \(uv\) using the formula
\begin{equation}
\frac{d(uv)}{dx} = \lim_{\Delta x \to 0} \frac{\Delta (uv)}{\Delta x}\tag{3.3.2}
\end{equation}
What is \(\Delta (uv) \text{?}\)
First look at what \(\Delta u \) and \(\Delta v \) are: \(\Delta u = f(x+\Delta x) - f(x) \text{,}\) so \(f(x+\Delta x) = u+\Delta u \text{,}\) and likewise \(g(x+\Delta x) = v+\Delta v \text{.}\)
Next, \(\Delta (uv)\) is the change in the value of the product \(f(x)g(x)\) as the argument changes from \(x \) to \(x+\Delta x\text{:}\)
\begin{equation*}
\Delta (uv) = f(x+\Delta x)g(x+\Delta x) - f(x)g(x)
= (u+\Delta u)(v+\Delta v) - uv
= u \Delta v + v \Delta u + \Delta u \Delta v
\end{equation*}
That is,
\begin{equation*}
\Delta (uv) = u \Delta v + v \Delta u + \Delta u \Delta v
\end{equation*}
The difference quotient in (3.3.2) is thus
\begin{equation}
\frac{\Delta (uv)}{\Delta x}
= \frac{u \Delta v + v \Delta u + \Delta u \Delta v}{\Delta x}
= \frac{u \Delta v}{\Delta x}+\frac{v \Delta u}{\Delta x}+\frac{\Delta u \Delta v}{\Delta x}
= u \frac{\Delta v}{\Delta x}+v \frac{\Delta u}{\Delta x} + \frac{\Delta u}{\Delta x} \frac{\Delta v}{\Delta x} \Delta x\tag{3.3.3}
\end{equation}
The two difference quotients here have limits as \(\Delta x \to 0\text{,}\) as do the factors multiplying them: \(\displaystyle \frac{\Delta u}{\Delta x} \to \frac{du}{dx} \text{,}\) \(\displaystyle \frac{\Delta v}{\Delta x} \to \frac{dv}{dx}\text{,}\) \(\Delta x \to 0\text{,}\) and \(u\) and \(v\) do not vary as \(\Delta x \to 0\text{.}\) Thus we can compute the limit as \(\Delta x \to 0\) in Equation (3.3.3):
Theorem 3.3.18. 1. The Product Rule for Derivatives.
\begin{equation*}
\frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}, \text{ or } (fg)' = f' g + f g'.
\end{equation*}
Example 3.3.19.
- If \(f(x)=xe^x\text{,}\) calculate its derivative \(f'(x)\text{.}\)
- Compute the second and third derivatives \(f''\) and \(f'''\) of this function.
- Compute all the derivatives \(f^{(n)}\) of this function.
Example 3.3.20.
Differentiate (compute the derivative of) the function \(f(t)=\sqrt{t}(a+bt).\) Do this two ways, with and without the Product Rule. Hint: as usual, it can help to rewrite roots as powers, and division by powers and roots as negative powers.
Sometimes, you only know the values of a function and its derivative at one point, like having measurements of the position and velocity of an object at one time. This can be enough to compute the derivative of another function got from the first with a product or such:
Example 3.3.21.
If \(f(x)=\sqrt{x}g(x)\text{,}\) calculate \(f'(x)\) in terms of \(x\) and \(g'(x)\text{,}\) and use this to find \(f'(4)\) given that \(g(4)=2\) and \(g'(4)=3\text{.}\)
The Derivative of the Reciprocal of a Function.
To deal with division, start with the simplest case, \(f(x) = 1/g(x)\text{,}\) and use a common strategy:
Rephrase a new problem in terms of a problem we have already solved.
In this case, we can restate the situation in terms of a product, and use the Product Rule. First, clear the denominator, getting \(f(x) g(x) = 1\text{.}\) Then use the product rule to get
\begin{equation*}
f'(x) g(x) + f(x) g'(x) = 0.
\end{equation*}
Solve for \(f'\) and substitute in \(f(x) = 1/g(x)\text{:}\)
\begin{equation*}
f'(x) = -g'(x) f(x)/g(x) = -g'(x)/[g(x)]^2,
\end{equation*}
or
\begin{equation*}
\left(\frac{1}{g}\right)' = -\frac{g'}{g^2}.
\end{equation*}
The minus sign goes with the fact that an increase in \(g\) will cause a decrease in \(1/g\text{.}\)
The Quotient Rule for Derivatives.
Now it is east to get a rule for the derivative of any quotient, by combining the product and reciprocal rules:
\begin{equation*}
\left( \frac{f}{g} \right)' = \left( f \cdot \frac{1}{g} \right)' = f' \cdot \frac{1}{g} + f \cdot \left(\frac{1}{g}\right)' = \frac{f'}{g} - f \cdot \frac{g'}{g^2} = \frac{f' g - f g'}{g^2}.
\end{equation*}
Theorem 3.3.22. The Quotient Rule for Derivatives.
\begin{equation*}
\left(\frac{f}{g}\right)' = \frac{f' g - f g'}{g^2}, \quad \text{or} \quad
\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{\displaystyle \frac{d}{dx}[f(x)] \cdot g(x) - f(x) \cdot \frac{d}{dx}[g(x)]}{[g(x)]^2}
\end{equation*}
Example 3.3.23.
- Differentiate \(\displaystyle y= \frac{x^2+x-2}{x^3+6}.\)
- Find the equation of the tangent line to this curve at point \(P(-1,-2/5).\)
Remark 3.3.24.
The derivatives of exponential functions are not covered in the OpenStax text till Section 3.9, but these functions are so important that I like to introduce these facts as soon as possible.Derivative of the Natural Exponential Function.
One way to define the number \(e\) is so that the slope of \(y=e^x\) at point \((0,1)\) is 1. That is,
\begin{equation*}
\lim_{h \to 0}\frac{e^{h}-1}{h} = 1.
\end{equation*}
This choice makes the derivative of \(e^x\) simple:
\begin{align*}
\frac{d}{dx}e^x
\amp = \lim_{h \to 0}\frac{e^{x+h}-e^{x}}{h}
= \lim_{h \to 0}\frac{e^x \cdot e^h - e^x}{h}
= \lim_{h \to 0}\frac{e^x (e^h - 1)}{h}\\
\amp = e^x \lim_{h \to 0}\frac{e^h - 1}{h}\\
\amp \text{(as } e^x \text{ is a constant as far as this limit is concerned: } h \text{ is the variable!)}\\
\amp = e^x \cdot 1\\
\amp = e^x.
\end{align*}
So the natural exponential function is equal to its own derivative:
Theorem 3.3.25. Derivatives of Exponential Functions.
\begin{equation*}
\frac{d}{dx}e^x = e^x.
\end{equation*}
Further, as will be shown in Section 3.4,
\begin{equation*}
\frac{d}{dx}a^x = (\ln a) a^x
\quad \text{for any constant } a > 0.
\end{equation*}
Thus all exponential functions have a rate of change proportional to their current value.
This fits for example with the simple exponential model of a population whose growth rate is proportional to its current size because the rates of births and deaths are both proportional to current population.
Geometrical Explanation of the Derivative of \(\mathbf{a^x}\).
The result for the derivative of \(a^x\) can be seen graphically by writing \(a=e^{\ln a}\) so that
\begin{equation*}
a^x=(e^{\ln a})^x = e^{(\ln a) x}.
\end{equation*}
The effect of changing from \(f(x)\) to \(g(x)=f(kx)\) is to compress the graph horizontally by a factor of \(k\text{,}\) increasing the slope at corresponding points by a factor \(k\text{:}\) in terms of derivatives,
\begin{equation*}
\frac{d}{dx} f(k x) = k f'(k x)
\end{equation*}
Thus, the graph of \(a^x\) is a compression of the graph of \(e^x\) by factor \(\ln a\) and
\begin{equation*}
\frac{d}{dx}a^x = (\ln a) e^{(\ln a) x} = (\ln a) a^x.\text{.}
\end{equation*}
In Section 3.4 we will see another way to compute this derivative, using a derivative rule for composition of functions. Example 3.3.26.
If \(f(x)=e^x - x\text{,}\) find \(f'(x)\) and \(f''(x)\text{,}\) and then compare the graphs of \(f\) and \(f''\text{.}\)
Answer 1.\(f'(x) = e^x - 1\)
\(f''(x) = e^x\)
Checkpoint 3.3.27.
At what point on the curve \(y=e^x\) is the tangent parallel to the line \(y=2x\text{?}\)
Differentiation Facts So Far.
- \(\displaystyle \displaystyle \frac{d}{dx}(c) = 0\)
- \(\displaystyle \displaystyle \frac{d}{dx}(x^n) = n x^{n-1}\)
- \(\displaystyle \displaystyle \frac{d}{dx}(e^x) = e^x\)
- \(\displaystyle \displaystyle \frac{d}{dx}(a^x) = (\ln a)a^x\)
- \(\displaystyle \displaystyle \frac{d}{dx}(cf) = c \frac{df}{dx}\)
- \(\displaystyle \displaystyle \frac{d}{dx}(f+g) = \frac{df}{dx} + \frac{dg}{dx}\)
- \(\displaystyle \displaystyle \frac{d}{dx}(f-g) = \frac{df}{dx} - \frac{dg}{dx}\)
- \(\displaystyle \displaystyle \frac{d}{dx}(fg) = \frac{df}{dx}g + f \frac{dg}{dx}\)
- \(\displaystyle \displaystyle \frac{d}{dx}\left(\frac{f}{g}\right) = \frac{\displaystyle\frac{df}{dx}g - f \frac{dg}{dx}}{g^2}\)
What is missing so far? Mostly, derivatives of
- trigonometric functions,
- inverses of functions, including logarithms, and
- compositions of functions.
These gaps will be filled in the next few sections.
Exercises Exercises
Study Calculus Volume 1, Section 3.3 3 ; in particular all Examples are worth reviewing, along with Checkpoint items 12 to 19 and Exercises 107, 109, 111, 119, 122, 127, 129, 130, 131, 133, 142, 143 and 147.
openstax.org/books/calculus-volume-1/pages/3-3-differentiation-rulesopenstax.org/books/calculus-volume-1/pages/3-9-derivatives-of-exponential-and-logarithmic-functionsopenstax.org/books/calculus-volume-1/pages/3-3-differentiation-rules
