4. The derivative¶

Chapter 1 found the speed of a falling stone at one instant by taking averages over shrinking intervals. Chapter 2 made "shrinking" precise. Now we put the two together and give the result its name: the derivative. In this chapter you will define the derivative at a point and as a function, draw tangent lines, see why a function with a derivative cannot jump but a function without jumps can still fail to have a derivative, and prove the rules (sum, product, quotient, power) that let you differentiate any polynomial or rational function in a line, with no limits in sight.

In this chapter

You will learn:

the definition of the derivative as a limit of difference quotients, and the tangent line it gives
the derivative as a new function, its notations, and higher derivatives such as acceleration
that differentiable functions are continuous (with proof), and the ways differentiability can fail: corners, vertical tangents, oscillation
the sum, constant multiple, product and quotient rules, with proofs
the power rule for whole-number, negative and rational exponents, with proofs

You need: Chapter 1 (rates as limits of averages), Chapter 2 (limits and the limit laws) and Chapter 3 (continuity, and in 4.5 the continuity of roots). Gentler versions on the Topics shelf: From secant to tangent, The derivative from first principles, The power rule, The derivative as a function, Velocity and acceleration and The product, quotient and chain rules.

Levels: 4.1 and 4.2 are intermediate; 4.3 is intermediate with an advanced ending; 4.4 is intermediate; 4.5 is intermediate to advanced.

4.1 Definition and tangent lines¶

The question¶

A curve is not a straight line, so it does not have one slope. But zoom in on a smooth curve near a point and it looks more and more like a straight line. What is the slope of that line? This is the tangent question from Chapter 1, and it is the same question as "what is the speed at an instant?" when the curve is a distance-time graph.

The method is the one we used for the stone. To find the slope of the graph of \(f\) at the point \((a, f(a))\), take a second point \((a + h, f(a + h))\) on the graph, with \(h \ne 0\). The straight line through the two points, the secant, has slope

\[ \frac{f(a + h) - f(a)}{h}, \]

the change in output divided by the change in input. This fraction is called a difference quotient. Now let \(h\) shrink to 0.

Definition. The derivative of \(f\) at \(a\) is

\(f'(a) = \lim_{h \to 0} \dfrac{f(a + h) - f(a)}{h}\),

if this limit exists (as a finite number). Then \(f\) is said to be differentiable at \(a\).

\(f'(a)\) is read "f prime of a". Putting \(x = a + h\) gives an equivalent form that is sometimes handier:

\[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}. \]

The tangent line to the graph at \((a, f(a))\) is the line through that point with slope \(f'(a)\):

\[ y = f(a) + f'(a)(x - a). \]

Worked examples from the definition¶

Example 1. \(f(x) = x^2\) at \(a = 3\).

\[ \frac{(3 + h)^2 - 9}{h} = \frac{6h + h^2}{h} = 6 + h \to 6 . \]

So \(f'(3) = 6\). The tangent line at \((3, 9)\) is \(y = 9 + 6(x - 3)\), that is, \(y = 6x - 9\). Check: the line meets the parabola where \(x^2 = 6x - 9\), i.e. \((x - 3)^2 = 0\): only at \(x = 3\), a double root. That is what touching looks like algebraically.

Example 2. \(f(x) = \frac1x\) at \(a = 2\).

\[ \frac{\frac1{2 + h} - \frac12}{h} = \frac{2 - (2 + h)}{2h(2 + h)} = \frac{-1}{2(2 + h)} \to -\frac14 . \]

The tangent line at \((2, \frac12)\) is \(y = \frac12 - \frac14(x - 2) = 1 - \frac x4\). The slope is negative because \(\frac1x\) is falling. Check: \(f(2.01) - f(2) \approx -0.0024876\), and \(-\frac14 \times 0.01 = -0.0025\).

Example 3. \(f(x) = \sqrt x\) at \(a = 4\). Use the conjugate:

\[ \frac{\sqrt{4 + h} - 2}{h} = \frac{(4 + h) - 4}{h(\sqrt{4 + h} + 2)} = \frac{1}{\sqrt{4 + h} + 2} \to \frac14 . \]

The tangent line at \((4, 2)\) is \(y = 2 + \frac14(x - 4) = \frac x4 + 1\). This gives a quick estimate: \(\sqrt{4.1} \approx \frac{4.1}4 + 1 = 2.025\). The true value is \(2.02485\ldots\) (Chapter 7 makes this kind of linear approximation systematic.)

Example 4 (falling on Mars). A stone dropped on Mars has fallen \(s(t) = 1.855\,t^2\) metres after \(t\) seconds. Its velocity at time \(a\) is

\[ s'(a) = \lim_{h \to 0} \frac{1.855\big((a + h)^2 - a^2\big)}{h} = \lim_{h \to 0} 1.855(2a + h) = 3.71a \text{ m/s}. \]

Units matter: a derivative has the units of the output divided by the units of the input. Distance in metres over time in seconds gives metres per second.

What the derivative means¶

The derivative is the instantaneous rate of change of \(f\) at \(a\): how fast the output is changing compared with the input, at that moment. Some readings:

position \(s(t)\): \(s'(t)\) is velocity;
volume of water in a tank \(V(t)\): \(V'(t)\) is the flow rate in litres per minute;
cost \(C(q)\) of making \(q\) items: \(C'(q)\) is the marginal cost, roughly the cost of one more item;
a population \(P(t)\): \(P'(t)\) is the growth rate in individuals per year.

And geometrically, it is the slope of the tangent. Near \(a\),

\[ f(a + h) \approx f(a) + f'(a)\,h , \]

with an error that is small even compared with \(h\): the error divided by \(h\) is \(\frac{f(a+h) - f(a)}{h} - f'(a)\), which tends to 0. This is the precise sense in which "the curve looks like its tangent line when you zoom in".

What a tangent is not¶

At school a tangent to a circle is "a line that touches the circle at one point". That description fails for other curves, which is why we define tangents by slopes.

The tangent to \(y = x^3\) at the origin is \(y = 0\) (the slope there is \(\lim_{h \to 0} \frac{h^3}{h} = 0\)). It crosses the curve, from below to above, at the point of tangency.
The tangent to \(y = x^3 - 3x\) at \((1, -2)\) is horizontal: \(y = -2\). It meets the curve again at \(x = -2\), since \((-2)^3 - 3(-2) = -2\).
The tangent to a straight line is the line itself, which meets it everywhere.

The right description is the zoom one: the tangent is the line that the curve resembles most closely near the point.

4.2 The derivative as a function¶

A new function from an old one¶

Example 1 found \(f'(3) = 6\) for \(f(x) = x^2\). Doing it at every point at once:

\[ \frac{(x + h)^2 - x^2}{h} = 2x + h \to 2x . \]

So \(f'(x) = 2x\) for every \(x\). The derivative is not just a number at one point; letting the point vary gives a new function \(f'\), the derivative of \(f\), whose value at each \(x\) is the slope of \(f\) there. Its domain is the set of points where \(f\) is differentiable.

Worked examples. From the definition:

\(f(x) = x^3\): \(\frac{(x + h)^3 - x^3}{h} = 3x^2 + 3xh + h^2 \to 3x^2\), so \(f'(x) = 3x^2\).
\(f(x) = \frac1x\) (\(x \ne 0\)): \(\frac{\frac1{x + h} - \frac1x}{h} = \frac{-1}{x(x + h)} \to -\frac1{x^2}\), so \(f'(x) = -\frac1{x^2}\).
\(f(x) = \sqrt x\) (\(x > 0\)): the conjugate gives \(\frac1{\sqrt{x + h} + \sqrt x} \to \frac1{2\sqrt x}\), so \(f'(x) = \frac1{2\sqrt x}\). The function is defined at 0 but the derivative is not: at \(x = 0\) the difference quotient is \(\frac{\sqrt h}{h} = \frac1{\sqrt h} \to \infty\). The graph has a vertical tangent there.
\(f(x) = mx + c\) (a line): the difference quotient is \(m\) for every \(h\), so \(f'(x) = m\). A line's slope is its slope everywhere. In particular a constant function has derivative 0.

Look at the pattern in \(x^2 \to 2x\), \(x^3 \to 3x^2\), \(x^{-1} \to -x^{-2}\), \(x^{1/2} \to \frac12 x^{-1/2}\). It is the power rule \(x^n \to n x^{n-1}\), which section 4.5 proves.

Notation¶

Several notations are in use, each convenient for something.

notation	read	who and why
\(f'(x)\)	"f prime of x"	Lagrange (1797); short, good for naming functions
\(\dfrac{dy}{dx}\), \(\dfrac{df}{dx}\)	"d y by d x"	Leibniz (1670s); recalls \(\frac{\Delta y}{\Delta x}\), the ratio of small changes
\(\dfrac{d}{dx} f(x)\)	"d by d x of f of x"	Leibniz's notation used as an operation on a formula
\(\dot y\)	"y dot"	Newton; used in physics for derivatives with respect to time
\(Df\)	"D f"	the derivative as an operator that turns functions into functions

So "\(\frac{d}{dx} x^3 = 3x^2\)", "if \(y = x^3\) then \(\frac{dy}{dx} = 3x^2\)" and "if \(f(x) = x^3\) then \(f'(x) = 3x^2\)" all say the same thing. To evaluate a Leibniz derivative at a point, write \(\left.\frac{dy}{dx}\right|_{x = 2} = 12\).

Leibniz's \(\frac{dy}{dx}\) is not a fraction (it is the limit of the fraction \(\frac{\Delta y}{\Delta x}\), where \(\Delta\), capital delta, means "change in"). But it often behaves like one, which is why it survives: Chapter 5's chain rule looks like \(\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}\).

Reading a derivative from a graph¶

Since \(f'(x)\) is the slope of \(f\) at \(x\), you can sketch \(f'\) from a picture of \(f\):

where \(f\) is rising, \(f' > 0\); where \(f\) is falling, \(f' < 0\);
at a smooth peak or valley, the tangent is horizontal, so \(f' = 0\);
where \(f\) is steep, \(|f'|\) is large.

(The first two statements are made rigorous by the mean value theorem in Chapter 6. Here they are guided by the picture.)

Worked example. \(f(x) = x^3 - 3x\). Using the derivatives above (and the sum rule of 4.4), \(f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)\). Its sign:

\(x\)	\(x < -1\)	\(-1\)	\(-1 < x < 1\)	\(1\)	\(x > 1\)
\(f'(x)\)	\(+\)	0	\(-\)	0	\(+\)
\(f\)	rising	peak, \(f = 2\)	falling	valley, \(f = -2\)	rising

The picture shows exactly this: the orange cubic rises to a peak at \((-1, 2)\), falls to a valley at \((1, -2)\), then rises; the blue parabola \(f'\) is positive, zero, negative, zero, positive in step.

Higher derivatives¶

The derivative of \(f'\) is the second derivative \(f''\), also written \(\frac{d^2y}{dx^2}\); then \(f'''\), and \(f^{(n)}\) for the \(n\)th. For motion, the derivative of velocity is acceleration:

\[ s(t) = 1.855\,t^2, \qquad v(t) = s'(t) = 3.71\,t, \qquad a(t) = v'(t) = s''(t) = 3.71 . \]

The stone on Mars accelerates at a constant \(3.71\) m/s\(^2\), which is where we started: the fall law was built from that number. For \(f(x) = x^3 - 3x\): \(f'(x) = 3x^2 - 3\), \(f''(x) = 6x\), \(f'''(x) = 6\), and every further derivative is 0. The second derivative measures how the slope itself changes: the bending of the graph (Chapter 7 calls it concavity).

Worked example (a stone thrown up on Mars). Throw a stone straight up at 20 m/s on Mars. Its height after \(t\) seconds is \(h(t) = 20t - 1.855t^2\) (ignoring the thin air). Its velocity is \(h'(t) = 20 - 3.71t\): positive while rising, zero at the top, negative on the way down. The top is where \(h'(t) = 0\): \(t = \frac{20}{3.71} \approx 5.39\) s, and the height there is \(h(5.39) \approx 53.9\) m. On Earth (replace \(1.855\) by \(4.9\)) the same throw tops out after about 2.04 s at about 20.4 m. Check: in both cases the top height is \(\frac{v_0^2}{2g}\): \(\frac{400}{7.42} \approx 53.9\) and \(\frac{400}{19.6} \approx 20.4\).

4.3 Differentiable implies continuous¶

The theorem¶

A function with a derivative at a point has a well-defined tangent there. Intuitively it cannot also have a jump: near a jump the secants become vertical. The theorem:

Theorem. If \(f\) is differentiable at \(a\), then \(f\) is continuous at \(a\).

Proof. For \(x \ne a\) we can write

\[ f(x) - f(a) = \frac{f(x) - f(a)}{x - a} \cdot (x - a). \]

As \(x \to a\), the first factor tends to \(f'(a)\) (a finite number, by assumption) and the second tends to 0. By the product law for limits, \(f(x) - f(a) \to f'(a) \cdot 0 = 0\). So \(\lim_{x \to a} f(x) = f(a)\). \(\square\)

The proof is three lines, but it rests on the word "finite". If the difference quotients blew up, the product \(\infty \cdot 0\) would tell us nothing.

The converse is false: |x|¶

Continuity does not imply differentiability. The standard example is the absolute value \(f(x) = |x|\) at \(a = 0\). It is continuous (Chapter 3.1). Its difference quotient at 0 is

\[ \frac{|0 + h| - |0|}{h} = \frac{|h|}{h} = \begin{cases} 1, & h > 0, \\ -1, & h < 0. \end{cases} \]

The right-hand limit is 1 and the left-hand limit is \(-1\). They differ, so \(f'(0)\) does not exist. These one-sided limits are called the right and left derivatives; \(f\) is differentiable at \(a\) exactly when both exist and agree. The graph of \(|x|\) has a corner at 0: from the right it looks like a line of slope 1, from the left like a line of slope \(-1\), and no single line fits both. Elsewhere \(|x|\) is differentiable, with derivative \(1\) for \(x > 0\) and \(-1\) for \(x < 0\): that is, \(\frac{d}{dx}|x| = \operatorname{sgn}(x)\) for \(x \ne 0\).

Other ways to fail¶

A vertical tangent. \(f(x) = x^{1/3}\) (the cube root) is continuous everywhere, but at 0 the difference quotient is \(\frac{h^{1/3}}{h} = h^{-2/3} \to \infty\). The tangent at the origin is the vertical line \(x = 0\), which has no slope. Similarly \(\sqrt x\) at 0 (from the right).
A cusp. \(f(x) = |x|^{1/2}\): the right derivative is \(+\infty\) and the left is \(-\infty\); the graph comes to a sharp point.
Oscillation. \(f(x) = x\sin\frac1x\) with \(f(0) = 0\) is continuous at 0 (Chapter 3, exercise 3). Its difference quotient at 0 is \(\frac{h \sin(1/h)}{h} = \sin\frac1h\), which has no limit (Chapter 2). So it is not differentiable at 0: the secants from the origin swing back and forth between slopes \(-1\) and \(1\) forever.
A jump. Any discontinuity at all, by the theorem.

Subtler: a derivative that is not continuous¶

Level: advanced.

Multiply by one more \(x\): let \(g(x) = x^2\sin\frac1x\) for \(x \ne 0\) and \(g(0) = 0\). At 0,

\[ \left|\frac{g(h) - g(0)}{h}\right| = \left|h \sin\frac1h\right| \le |h| \to 0, \]

so by the squeeze theorem \(g'(0) = 0\). The function is differentiable at 0 even though it wiggles infinitely often nearby; the wiggles are squashed by the \(x^2\). For \(x \ne 0\), the rules of Chapter 5 give \(g'(x) = 2x\sin\frac1x - \cos\frac1x\), and as \(x \to 0\) the first term tends to 0 while \(\cos\frac1x\) oscillates between \(-1\) and \(1\). So \(g'\) exists everywhere but is not continuous at 0. Differentiable functions can have wild derivatives. (They cannot be completely wild: Darboux proved in 1875 that every derivative has the intermediate value property, even when it is not continuous.)

How bad can it get? In 1872 Weierstrass presented a function that is continuous at every point and differentiable at none: a graph that is all corners, at every scale. Chapter 19 builds it. So the hierarchy is strict: differentiable functions are continuous, but continuous functions can be very far from differentiable.

4.4 The rules: sum, product, quotient¶

Computing every derivative from the definition is slow. The rules below let you build derivatives of complicated functions from simple ones, the way the limit laws built limits. Throughout, \(f\) and \(g\) are differentiable at \(x\).

Constant multiples and sums¶

Constant multiple rule. \((cf)'(x) = c\,f'(x)\). Sum rule. \((f + g)'(x) = f'(x) + g'(x)\).

Proof. The difference quotient of \(f + g\) is

\[ \frac{f(x + h) + g(x + h) - f(x) - g(x)}{h} = \frac{f(x + h) - f(x)}{h} + \frac{g(x + h) - g(x)}{h}, \]

and by the sum law for limits it tends to \(f'(x) + g'(x)\). The constant multiple rule is the same with the constant multiple law. \(\square\)

So derivatives of sums are sums of derivatives (the derivative is linear). Then \(\frac{d}{dx}(4x^3 - 5x^2 + 7) = 12x^2 - 10x\), using \(\frac{d}{dx}x^3 = 3x^2\), \(\frac{d}{dx}x^2 = 2x\) and the constant's derivative 0.

The product rule¶

It is tempting to guess that the derivative of a product is the product of the derivatives. Test it: \(x \cdot x = x^2\) has derivative \(2x\), but the product of the derivatives is \(1 \cdot 1 = 1\). Wrong. The correct rule has two terms.

Product rule. \((fg)'(x) = f'(x)g(x) + f(x)g'(x)\).

Why two terms: a picture. Think of \(f(x)\) and \(g(x)\) as the sides of a rectangle with area \(fg\). When \(x\) grows a little, \(f\) grows by \(\Delta f\) and \(g\) by \(\Delta g\). The new area is

\[ (f + \Delta f)(g + \Delta g) = fg + \Delta f \cdot g + f \cdot \Delta g + \Delta f \cdot \Delta g . \]

The area grows by two thin strips, \(\Delta f \cdot g\) and \(f \cdot \Delta g\), plus a tiny corner \(\Delta f\,\Delta g\). Divide by \(\Delta x\) and let it shrink: the strips give \(f'g + fg'\), and the corner disappears because it is a product of two small quantities. Leibniz wrote it as \(d(xy) = x\,dy + y\,dx\), in his first publication on calculus (1684).

Proof. Add and subtract \(f(x + h)g(x)\) in the numerator (the same trick as the product law for limits):

\[ \begin{aligned} \frac{f(x + h)g(x + h) - f(x)g(x)}{h} &= \frac{f(x + h)g(x + h) - f(x + h)g(x)}{h} + \frac{f(x + h)g(x) - f(x)g(x)}{h} \\ &= f(x + h)\,\frac{g(x + h) - g(x)}{h} + g(x)\,\frac{f(x + h) - f(x)}{h}. \end{aligned} \]

As \(h \to 0\): \(f(x + h) \to f(x)\), because \(f\) is differentiable, hence continuous (section 4.3); the first difference quotient tends to \(g'(x)\), and the second to \(f'(x)\). By the limit laws the whole expression tends to \(f(x)g'(x) + g(x)f'(x)\). \(\square\)

Notice how the theorem of 4.3 was needed. That is typical: continuity is the glue that holds the rules together.

Worked example. \(\frac{d}{dx}\big[(x^2 + 1)(x^3 - 2x)\big] = 2x(x^3 - 2x) + (x^2 + 1)(3x^2 - 2) = 5x^4 - 3x^2 - 2\). Check: multiplying out first, \((x^2 + 1)(x^3 - 2x) = x^5 - x^3 - 2x\), whose derivative is \(5x^4 - 3x^2 - 2\). Same answer.

Worked example. With three factors, apply the rule twice: \((fgh)' = f'gh + fg'h + fgh'\). Each term differentiates one factor and leaves the others alone. For example \(\frac{d}{dx}\big[x\sqrt x\,(x + 1)\big] = \sqrt x\,(x + 1) + x \cdot \frac1{2\sqrt x}(x + 1) + x\sqrt x\), which simplifies to \(\frac52 x^{3/2} + \frac32 x^{1/2}\). Check: \(x\sqrt x(x + 1) = x^{5/2} + x^{3/2}\), and the power rule of 4.5 gives the same.

The quotient rule¶

Reciprocal rule. If \(g(x) \ne 0\), then \(\left(\dfrac1g\right)'(x) = -\dfrac{g'(x)}{g(x)^2}\).

Quotient rule. If \(g(x) \ne 0\), then \(\left(\dfrac fg\right)'(x) = \dfrac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\).

Proof of the reciprocal rule. Since \(g\) is continuous at \(x\) and \(g(x) \ne 0\), \(g(x + h) \ne 0\) for small \(h\) (Chapter 2, exercise 11), so \(\frac1{g(x + h)}\) makes sense. Then

\[ \frac{\frac{1}{g(x + h)} - \frac1{g(x)}}{h} = \frac{g(x) - g(x + h)}{h\,g(x + h)g(x)} = -\frac{g(x + h) - g(x)}{h} \cdot \frac{1}{g(x + h)g(x)} . \]

As \(h \to 0\) the first factor tends to \(-g'(x)\) and the second, by continuity of \(g\), to \(\frac1{g(x)^2}\). \(\square\)

Proof of the quotient rule. Write \(\frac fg = f \cdot \frac1g\) and use the product rule:

\[ \left(f \cdot \frac1g\right)' = f' \cdot \frac1g + f \cdot \left(-\frac{g'}{g^2}\right) = \frac{f'g - fg'}{g^2}. \qquad \square \]

A way to remember the order of the numerator: it is the product rule with a minus sign, and the term with \(f'\) comes first ("low d-high minus high d-low, over low squared", if you like mnemonics). If you get the order wrong, the sign is wrong; check with \(\frac{x}{1}\), whose derivative must be \(1\).

Worked example. \(\frac{d}{dx} \frac{x^2}{x + 1} = \frac{2x(x + 1) - x^2 \cdot 1}{(x + 1)^2} = \frac{x^2 + 2x}{(x + 1)^2}\). At \(x = 1\) this is \(\frac34\). Check numerically: \(\frac{f(1.001) - f(0.999)}{0.002}\) with \(f(x) = \frac{x^2}{x+1}\) is about \(0.75000\).

Worked example. \(\frac{d}{dx} \frac{3x - 1}{x^2 + 1} = \frac{3(x^2 + 1) - (3x - 1)(2x)}{(x^2 + 1)^2} = \frac{-3x^2 + 2x + 3}{(x^2 + 1)^2}\). The tangent is horizontal where \(3x^2 - 2x - 3 = 0\), at \(x = \frac{1 \pm \sqrt{10}}{3}\), that is \(x \approx 1.387\) and \(x \approx -0.721\).

Rational functions. By the quotient rule, every rational function is differentiable at every point of its domain, and its derivative is again a rational function.

4.5 The power rule¶

Power rule. \(\dfrac{d}{dx} x^n = n x^{n-1}\).

We prove it in stages: first for positive whole numbers \(n\), then for negative whole numbers, then for rational exponents. For arbitrary real exponents, such as \(x^{\sqrt2}\) or \(x^\pi\), we first need to say what such a power means, which needs the exponential function; Chapter 5 does that and proves the rule for all real \(n\) by logarithmic differentiation.

Positive whole numbers: three proofs¶

Proof 1 (binomial theorem). By the binomial theorem,

\[ (x + h)^n = x^n + n x^{n-1}h + \binom n2 x^{n-2}h^2 + \cdots + h^n . \]

Subtract \(x^n\) and divide by \(h\):

\[ \frac{(x + h)^n - x^n}{h} = n x^{n-1} + \binom n2 x^{n-2}h + \cdots + h^{n-1} . \]

Every term after the first contains a factor \(h\) and so tends to 0. The limit is \(nx^{n-1}\). \(\square\)

Proof 2 (induction with the product rule). For \(n = 1\), \(\frac{d}{dx}x = 1 = 1 \cdot x^0\). Suppose \(\frac{d}{dx} x^n = nx^{n-1}\) for some \(n\). Then by the product rule

\[ \frac{d}{dx} x^{n+1} = \frac{d}{dx}(x^n \cdot x) = nx^{n-1} \cdot x + x^n \cdot 1 = (n + 1)x^n , \]

which is the rule for \(n + 1\). By induction it holds for all \(n \ge 1\). \(\square\)

Proof 3 (factoring). For \(x \ne a\), the identity \(x^n - a^n = (x - a)(x^{n-1} + x^{n-2}a + \cdots + x a^{n-2} + a^{n-1})\) (multiply out the right side: everything cancels except \(x^n - a^n\)) gives

\[ \frac{x^n - a^n}{x - a} = x^{n-1} + x^{n-2}a + \cdots + a^{n-1} \to a^{n-1} + a^{n-1} + \cdots + a^{n-1} = n a^{n-1} \]

as \(x \to a\), since there are \(n\) terms, each tending to \(a^{n-1}\). \(\square\)

Proof 3 is the one that generalises, as you will see in a moment. With the sum and constant multiple rules, every polynomial can now be differentiated term by term:

\[ \frac{d}{dx}\left(c_n x^n + \cdots + c_1 x + c_0\right) = n c_n x^{n-1} + \cdots + c_1 . \]

Negative whole numbers¶

For \(n = -m\) with \(m\) a positive whole number and \(x \ne 0\), \(x^{-m} = \frac1{x^m}\), and the reciprocal rule gives

\[ \frac{d}{dx} x^{-m} = -\frac{m x^{m-1}}{x^{2m}} = -m x^{-m-1} , \]

which is \(n x^{n-1}\) with \(n = -m\). \(\square\) (And \(n = 0\): \(x^0 = 1\) has derivative \(0 = 0 \cdot x^{-1}\), for \(x \ne 0\).)

Worked example. \(\frac{d}{dx}\left(\frac{3}{x^4}\right) = 3 \cdot (-4)x^{-5} = -\frac{12}{x^5}\).

Rational exponents¶

Level: advanced.

Roots. Let \(n \ge 1\) be a whole number and \(x > 0\). We show \(\frac{d}{dx}x^{1/n} = \frac1n x^{\frac1n - 1}\).

Proof. Fix \(a > 0\) and write \(u = x^{1/n}\) and \(b = a^{1/n}\), so \(x = u^n\) and \(a = b^n\). For \(x \ne a\) (so \(u \ne b\)), by Proof 3's identity,

\[ \frac{x^{1/n} - a^{1/n}}{x - a} = \frac{u - b}{u^n - b^n} = \frac{1}{u^{n-1} + u^{n-2}b + \cdots + b^{n-1}} . \]

As \(x \to a\), \(u = x^{1/n} \to a^{1/n} = b\), because the \(n\)th root is continuous (Chapter 3.4: it is the inverse of the continuous increasing function \(x^n\)). So the denominator tends to \(nb^{n-1}\), which is positive, and the limit is

\[ \frac{1}{n b^{n-1}} = \frac1n a^{-\frac{n-1}{n}} = \frac1n a^{\frac1n - 1}. \qquad \square \]

Check. With \(n = 2\) this gives \(\frac{d}{dx}\sqrt x = \frac12 x^{-1/2} = \frac1{2\sqrt x}\), as found in 4.2.

Powers of a function. To reach \(x^{p/q} = \left(x^{1/q}\right)^p\) we need the derivative of a power of a function. For a whole number \(p \ge 1\) and \(u\) differentiable,

\[ \frac{d}{dx}\,u(x)^p = p\,u(x)^{p-1}\,u'(x). \]

Proof by induction on \(p\). For \(p = 1\) it says \(u' = u'\). If it holds for \(p\), then by the product rule \((u^{p+1})' = (u^p \cdot u)' = p u^{p-1}u' \cdot u + u^p \cdot u' = (p + 1)u^p u'\). \(\square\) (This is a special case of the chain rule of Chapter 5, proved here with the product rule alone.)

Positive rational exponents. For whole numbers \(p, q \ge 1\) and \(x > 0\), take \(u = x^{1/q}\):

\[ \frac{d}{dx}x^{p/q} = p\left(x^{1/q}\right)^{p-1} \cdot \frac1q x^{\frac1q - 1} = \frac pq x^{\frac{p-1}{q} + \frac1q - 1} = \frac pq x^{\frac pq - 1}. \]

Negative rational exponents follow from the reciprocal rule exactly as for whole numbers. So the power rule \(\frac{d}{dx}x^r = rx^{r-1}\) holds for every rational \(r\) and every \(x > 0\) (and at other \(x\) wherever \(x^r\) is defined and the formula makes sense, for example \(x^{1/3}\) at negative \(x\), by the oddness of the cube root).

Worked examples¶

Example A. \(\frac{d}{dx}\left(3x^4 - \frac{2}{x^2} + 5\sqrt x\right) = 12x^3 + \frac{4}{x^3} + \frac{5}{2\sqrt x}\).

Example B. \(\frac{d}{dx}\frac{x^2 + 1}{\sqrt x}\). Rather than the quotient rule, divide first: \(\frac{x^2 + 1}{\sqrt x} = x^{3/2} + x^{-1/2}\), so the derivative is \(\frac32 x^{1/2} - \frac12 x^{-3/2}\). Rewriting before differentiating often saves work.

Example C (Kepler's third law). For a planet in orbit, the period \(T\) (in years) and the average distance \(a\) from the Sun (in astronomical units, AU) satisfy Kepler's third law \(T = a^{3/2}\), in these units. Mars has \(a \approx 1.524\) AU, so \(T \approx 1.524^{3/2} \approx 1.881\) years, the Martian year. The derivative

\[ \frac{dT}{da} = \frac32 a^{1/2} \approx \frac32 \times 1.234 \approx 1.85 \text{ years per AU} \]

says that, near Mars's orbit, moving an orbit outward by 0.01 AU lengthens the year by about 0.0185 years, nearly a week. More neatly, \(\frac{dT}{da} = \frac32 \cdot \frac Ta\), so a 1 percent increase in distance gives about a 1.5 percent increase in period, at every orbit. (Chapter 14 derives Kepler's laws from Newton's.)

Example D. Where does the tangent to \(y = x^{2/3}\) have slope \(\frac13\)? \(y' = \frac23 x^{-1/3} = \frac13\) gives \(x^{1/3} = 2\), so \(x = 8\), at the point \((8, 4)\). At \(x = 0\) the derivative is undefined: \(x^{2/3}\) has a cusp there.

History¶

Pierre de Fermat, in the 1630s, found maxima, minima and tangents of polynomial curves with a method he called adequality: compare \(f(x)\) with \(f(x + e)\), divide by \(e\), then set \(e\) to zero. In modern terms he was computing derivatives of polynomials, but without a theory of why the method worked. Isaac Newton developed his "method of fluxions" from the mid-1660s; his main treatise on it was written around 1671 and published only in 1736, after his death. Gottfried Wilhelm Leibniz published first: his paper Nova Methodus pro Maximis et Minimis ("A new method for maxima and minima") appeared in the journal Acta Eruditorum in 1684, with the \(d\) notation, the sum, product and quotient rules and the power rule. Joseph-Louis Lagrange, in his Théorie des fonctions analytiques (1797), introduced the prime notation and the name "derived function" (fonction dérivée), from which our word "derivative" comes. The definition as a limit, with the rules proved as in this chapter, is Cauchy's, from the 1820s.

Check yourself¶

1. From the definition, find \(f'(1)\) for \(f(x) = x^2 + 3x\), and the tangent line at \(x = 1\).

\(\frac{(1 + h)^2 + 3(1 + h) - 4}{h} = \frac{5h + h^2}{h} = 5 + h \to 5\). Since \(f(1) = 4\), the tangent is \(y = 4 + 5(x - 1) = 5x - 1\). Check with the rules: \(f'(x) = 2x + 3\), \(f'(1) = 5\).

2. From the definition, find the derivative of \(f(x) = \dfrac1{x + 1}\).

\(\frac{1}{h}\left(\frac1{x + h + 1} - \frac1{x + 1}\right) = \frac{(x + 1) - (x + h + 1)}{h(x + h + 1)(x + 1)} = \frac{-1}{(x + h + 1)(x + 1)} \to -\frac{1}{(x + 1)^2}\), for \(x \ne -1\).

3. From the definition, find the derivative of \(f(x) = \sqrt{2x + 1}\) for \(x > -\tfrac12\).

Multiply by the conjugate: \(\frac{\sqrt{2x + 2h + 1} - \sqrt{2x + 1}}{h} = \frac{2h}{h(\sqrt{2x + 2h + 1} + \sqrt{2x + 1})} \to \frac{2}{2\sqrt{2x + 1}} = \frac1{\sqrt{2x + 1}}\).

4. Where is \(f(x) = |x - 2| + |x + 1|\) not differentiable? What are its one-sided derivatives there?

At \(x = 2\) and \(x = -1\), the corners of the two absolute values. Away from them: for \(x > 2\), \(f = 2x - 1\) (slope 2); for \(-1 < x < 2\), \(f = 3\) (slope 0); for \(x < -1\), \(f = -2x + 1\) (slope \(-2\)). At \(x = 2\) the left derivative is 0 and the right is 2; at \(x = -1\) the left is \(-2\) and the right is 0.

5. Is \(f(x) = x|x|\) differentiable at 0? Find \(f'(x)\) everywhere.

At 0: \(\frac{h|h|}{h} = |h| \to 0\), so \(f'(0) = 0\). For \(x > 0\), \(f = x^2\) and \(f' = 2x\); for \(x < 0\), \(f = -x^2\) and \(f' = -2x\). Together, \(f'(x) = 2|x|\) for all \(x\): differentiable everywhere, though \(f'\) itself has a corner at 0.

6. Differentiate \((x^2 + 1)(x^3 - 2x)\) with the product rule, and evaluate at \(x = 1\).

\(2x(x^3 - 2x) + (x^2 + 1)(3x^2 - 2)\). At \(x = 1\): \(2(-1) + 2(1) = 0\). So the tangent at \((1, -2)\) is horizontal. Check: expanding gives \(5x^4 - 3x^2 - 2\), and \(5 - 3 - 2 = 0\).

7. Differentiate \(\dfrac{x^2 - 1}{x^2 + 1}\) and find where the tangent is horizontal.

\(\frac{2x(x^2 + 1) - (x^2 - 1)2x}{(x^2 + 1)^2} = \frac{4x}{(x^2 + 1)^2}\). Zero only at \(x = 0\), the point \((0, -1)\), which is the lowest point of the graph (the derivative is negative before 0 and positive after).

8. Differentiate \(3x^4 - \dfrac{2}{x^2} + 5\sqrt x\) and \(x^{5/3} - 6x^{-1/3}\).

\(12x^3 + 4x^{-3} + \frac52 x^{-1/2}\), and \(\frac53 x^{2/3} + 2x^{-4/3}\) (since \(-6 \cdot (-\frac13) = 2\) and \(-\frac13 - 1 = -\frac43\)).

9. Find the points of \(y = x^3 - 3x\) where the tangent is horizontal, and the points where it is parallel to \(y = 9x\).

\(y' = 3x^2 - 3 = 0\) at \(x = \pm1\): points \((1, -2)\) and \((-1, 2)\). Parallel to \(y = 9x\): \(3x^2 - 3 = 9\), \(x = \pm 2\): points \((2, 2)\) and \((-2, -2)\).

10. Which tangent line to \(y = \sqrt x\) passes through the point \((-4, 0)\)?

The tangent at \((a, \sqrt a)\) is \(y = \sqrt a + \frac{x - a}{2\sqrt a}\). It passes through \((-4, 0)\) when \(0 = \sqrt a + \frac{-4 - a}{2\sqrt a}\), i.e. \(2a = 4 + a\), so \(a = 4\). The line is \(y = 2 + \frac{x - 4}4 = \frac x4 + 1\), and indeed it is 0 at \(x = -4\).

11. A stone is thrown straight up at 20 m/s on Mars, with height \(h(t) = 20t - 1.855t^2\). When is it at its highest, how high, and with what velocity does it land?

\(h'(t) = 20 - 3.71t = 0\) at \(t \approx 5.39\) s, height \(\approx 53.9\) m. It lands when \(h(t) = 0\), at \(t = \frac{20}{1.855} \approx 10.78\) s (twice the rise time), with velocity \(h'(10.78) = 20 - 3.71 \times 10.78 \approx -20\) m/s: the same speed it was thrown with, downward, as you would expect without air.

12. Let \(g(x) = x^2\sin\frac1x\) for \(x \ne 0\) and \(g(0) = 0\). Show \(g'(0) = 0\), but \(x\sin\frac1x\) (with value 0 at 0) has no derivative at 0. Then use the power-of-a-function rule to show \(\frac{d}{dx}(x^2 + 1)^{10} = 20x(x^2 + 1)^9\).

For \(g\): \(\left|\frac{g(h)}{h}\right| = |h\sin\frac1h| \le |h| \to 0\), so \(g'(0) = 0\) by the squeeze theorem. For \(f(x) = x\sin\frac1x\): \(\frac{f(h)}{h} = \sin\frac1h\) has no limit as \(h \to 0\) (Chapter 2, exercise 10), so \(f'(0)\) does not exist. Finally, with \(u = x^2 + 1\), \(u' = 2x\) and \(p = 10\): \(10u^9 \cdot 2x = 20x(x^2 + 1)^9\).

What you learned¶

The derivative \(f'(a) = \lim_{h \to 0}\frac{f(a + h) - f(a)}{h}\) is the instantaneous rate of change and the slope of the tangent line \(y = f(a) + f'(a)(x - a)\).
Letting \(a\) vary gives the derivative function \(f'\), written also \(\frac{dy}{dx}\), \(\dot y\) or \(Df\); its sign shows where \(f\) rises and falls, and \(f''\) gives acceleration and bending.
Differentiable implies continuous, because \(f(x) - f(a)\) is a difference quotient times \(x - a\).
Continuous does not imply differentiable: corners (\(|x|\)), vertical tangents (\(x^{1/3}\)), cusps and oscillation (\(x\sin\frac1x\)); and a derivative can exist everywhere without being continuous (\(x^2\sin\frac1x\)).
The derivative is linear; the product rule is \((fg)' = f'g + fg'\) and the quotient rule \(\left(\frac fg\right)' = \frac{f'g - fg'}{g^2}\), both proved using continuity.
The power rule \(\frac{d}{dx}x^r = rx^{r-1}\) holds for every rational \(r\) (for real \(r\) after Chapter 5), proved by the binomial theorem, induction or factoring.

Next: Chapter 5 proves the chain rule and differentiates the sine, cosine, exponential and logarithm, and inverse functions.