Unit 3, lesson 21

The Derivative as a Function

How the derivative becomes a new function, the notations used for it, what differentiability on an interval means, why it implies continuity, and how corners, cusps and vertical tangents break it.

  • Theory part 41
  • Practice part 42
  • Definition of the Derivative

Theory, part 41

Play the video

Nothing loads from YouTube until you press play.

Practice, part 42

Play the video

Nothing loads from YouTube until you press play.

Computing a derivative at x=1x = 1, then again at x=2x = 2, then at x=3x = 3 gets old fast. It is better to do the computation once with a general xx and get a formula that gives the slope at every point. That formula is a function in its own right, and studying where it exists tells you exactly where a graph is smooth.

Key terms

The derivative function

Definition For a function ff, the derivative of ff is the function f′f' given by

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}

at every xx for which ff is defined on an open interval around xx and this limit exists as a finite number. The domain of f′f' is the set of all such xx, which can be smaller than the domain of ff.

In plain words Plug in a general point instead of a specific one, and the derivative becomes a slope formula for the whole graph.

Example For f(x)=x3f(x) = x^3,

(x+h)3−x3h=3x2h+3xh2+h3h=3x2+3xh+h2→3x2,\frac{(x + h)^3 - x^3}{h} = \frac{3x^2 h + 3x h^2 + h^3}{h} = 3x^2 + 3xh + h^2 \to 3x^2,

so f′(x)=3x2f'(x) = 3x^2 for every real xx. For g(x)=xg(x) = \sqrt{x} the same method gives g′(x)=12xg'(x) = \frac{1}{2\sqrt{x}}, but only for x>0x > 0. The domain of gg is [0,∞)[0, \infty) while the domain of g′g' is (0,∞)(0, \infty).

Leibniz and prime notation

Definition If y=f(x)y = f(x), each of the following denotes the derivative.

f′(x),y′,dydx,dfdx,ddxf(x),Df(x).f'(x), \qquad y', \qquad \frac{dy}{dx}, \qquad \frac{df}{dx}, \qquad \frac{d}{dx} f(x), \qquad D f(x).

The value at a specific point aa is written f′(a)f'(a) or dydx∣x=a\left. \frac{dy}{dx} \right\vert_{x = a}. The symbol dydx\frac{dy}{dx} is a single symbol for a limit, not a fraction of two numbers, although it often behaves like one in calculations.

In plain words Prime notation is short. Leibniz notation shows which variable is changing, which helps when several variables are in play.

Example If y=x3y = x^3 then y′=3x2y' = 3x^2, dydx=3x2\frac{dy}{dx} = 3x^2 and dydx∣x=2=12\left. \frac{dy}{dx} \right\vert_{x = 2} = 12. All three say the same thing, and the last one is the slope at the point (2,8)(2, 8).

The differentiation operator

Definition The differentiation operator ddx\frac{d}{dx}, also written DD, takes a function ff and returns the function f′f', defined wherever ff is differentiable. It acts on whole functions, not on single numbers.

In plain words Think of ddx\frac{d}{dx} as a machine. A function goes in, its derivative function comes out.

Example Applying the operator to x3x^3 gives ddx(x3)=3x2\frac{d}{dx}\left(x^3\right) = 3x^2. Applying it again gives ddx(3x2)=6x\frac{d}{dx}\left(3x^2\right) = 6x, which can be checked by the same limit,

3(x+h)2−3x2h=6xh+3h2h=6x+3h→6x.\frac{3(x + h)^2 - 3x^2}{h} = \frac{6xh + 3h^2}{h} = 6x + 3h \to 6x.

Note that ddx(f(2))\frac{d}{dx}\left(f(2)\right) is the derivative of a constant, not f′(2)f'(2).

One-sided derivatives

Definition Suppose ff is defined on [a,a+r)[a, a + r) for some r>0r > 0. The right-hand derivative of ff at aa is

f+′(a)=lim⁡h→0+f(a+h)−f(a)h.f'_+(a) = \lim_{h \to 0^+} \frac{f(a + h) - f(a)}{h}.

Suppose ff is defined on (a−r,a](a - r, a] for some r>0r > 0. The left-hand derivative of ff at aa is

f−′(a)=lim⁡h→0−f(a+h)−f(a)h.f'_-(a) = \lim_{h \to 0^-} \frac{f(a + h) - f(a)}{h}.

Each one exists only when its limit exists as a finite number. For ff defined on an open interval around aa, the derivative f′(a)f'(a) exists exactly when both one-sided derivatives exist and are equal, and then f′(a)f'(a) is their common value.

In plain words Measure the slope using only points to the right, then only points to the left. A true derivative needs both answers to agree.

Example For f(x)=∣x∣f(x) = \lvert x \rvert at 00 the quotient is ∣h∣h\frac{\lvert h \rvert}{h}, which is 11 for h>0h > 0 and −1-1 for h<0h < 0. So f+′(0)=1f'_+(0) = 1 and f−′(0)=−1f'_-(0) = -1. They differ, so ∣x∣\lvert x \rvert is not differentiable at 00.

Differentiability on an interval

Definition A function ff is differentiable on an open interval (a,b)(a, b), which may be unbounded, if it is differentiable at every point of it. It is differentiable on a closed interval [a,b][a, b] if it is differentiable on (a,b)(a, b) and the one-sided derivatives f+′(a)f'_+(a) and f−′(b)f'_-(b) exist. Half-open intervals are handled the same way at the included endpoint.

In plain words Inside the interval use ordinary derivatives. At an included endpoint only the slope from inside is required.

Example The function x2x^2 is differentiable on [0,1][0, 1], with f+′(0)=0f'_+(0) = 0 and f−′(1)=2f'_-(1) = 2. The function x\sqrt{x} is differentiable on (0,1](0, 1] but not on [0,1][0, 1], because at 00 the quotient hh=1h\frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}} tends to ∞\infty as h→0+h \to 0^+, so f+′(0)f'_+(0) does not exist.

Differentiability implies continuity

Definition If ff is differentiable at aa, then ff is continuous at aa. Proof. For h≠0h \neq 0,

f(a+h)−f(a)=f(a+h)−f(a)h⋅h→f′(a)⋅0=0as h→0,f(a + h) - f(a) = \frac{f(a + h) - f(a)}{h} \cdot h \to f'(a) \cdot 0 = 0 \quad \text{as } h \to 0,

so lim⁡h→0f(a+h)=f(a)\lim_{h \to 0} f(a + h) = f(a). The converse is false, since a function can be continuous at a point without being differentiable there.

In plain words A graph that has a tangent line at a point cannot have a break there. But an unbroken graph can still fail to have a tangent.

Example The floor function jumps at x=1x = 1, so it is not continuous there, and therefore it cannot be differentiable at 11. In the other direction, ∣x∣\lvert x \rvert is continuous at 00 but not differentiable there.

Corners

Definition Let ff be defined on an open interval around aa. Then ff has a corner at aa if the one-sided derivatives f−′(a)f'_-(a) and f+′(a)f'_+(a) both exist as finite numbers and are different. Then ff is continuous at aa, because each one-sided derivative forces continuity from that side, but f′(a)f'(a) does not exist. Some books also call a point a corner when the slope is finite on one side and infinite on the other.

In plain words The graph arrives along one straight direction and leaves along another, making a sharp bend.

Example Let f(x)=∣x2−4∣f(x) = \lvert x^2 - 4 \rvert at a=2a = 2, where f(2)=0f(2) = 0. For small h>0h > 0 we have f(2+h)=(2+h)2−4f(2 + h) = (2 + h)^2 - 4, and for small h<0h < 0 we have f(2+h)=4−(2+h)2f(2 + h) = 4 - (2 + h)^2. So the quotients are

f(2+h)−f(2)h={4+h→4,h→0+,−4−h→−4,h→0−.\frac{f(2 + h) - f(2)}{h} = \begin{cases} 4 + h \to 4, & h \to 0^+, \\ -4 - h \to -4, & h \to 0^-. \end{cases}

So f+′(2)=4f'_+(2) = 4, f−′(2)=−4f'_-(2) = -4, and the graph has a corner at x=2x = 2.

Cusps

Definition Let ff be defined on an open interval around aa and continuous at aa. Then ff has a cusp at aa if the difference quotient tends to ∞\infty from one side of aa and to −∞-\infty from the other side. That is, one of the limits

lim⁡h→0−f(a+h)−f(a)handlim⁡h→0+f(a+h)−f(a)h\lim_{h \to 0^-} \frac{f(a + h) - f(a)}{h} \quad \text{and} \quad \lim_{h \to 0^+} \frac{f(a + h) - f(a)}{h}

is ∞\infty and the other is −∞-\infty.

In plain words Both sides of the graph become vertical as they meet, but one comes down and the other goes up, forming a sharp point. Continuity matters here, since without it a jump could also make the quotients blow up.

Example For f(x)=x2/3f(x) = x^{2/3} at 00, the quotient is h2/3h=h−1/3\frac{h^{2/3}}{h} = h^{-1/3}. As h→0+h \to 0^+ this tends to ∞\infty, and as h→0−h \to 0^- it tends to −∞-\infty, since the cube root of a small negative number is a small negative number. The graph has a cusp at the origin.

Vertical tangents

Definition Let ff be defined on an open interval around aa and continuous at aa. Then ff has a vertical tangent at aa if the difference quotient tends to ∞\infty from both sides of aa, or to −∞-\infty from both sides. The tangent line is then the vertical line x=ax = a, and f′(a)f'(a) does not exist.

In plain words The curve passes straight up or straight down for an instant. A vertical line has no slope, so there is no derivative.

Example For f(x)=x−23f(x) = \sqrt[3]{x - 2} at a=2a = 2 the quotient is

f(2+h)−f(2)h=h3h=1h23,\frac{f(2 + h) - f(2)}{h} = \frac{\sqrt[3]{h}}{h} = \frac{1}{\sqrt[3]{h^2}},

which is positive and tends to ∞\infty as h→0h \to 0 from either side. So the graph has the vertical tangent x=2x = 2 at the point (2,0)(2, 0).

Check yourself

  1. Use the limit definition to find f′(x)f'(x) for f(x)=x2−5xf(x) = x^2 - 5x.
Show the answer

The difference quotient simplifies to

(x+h)2−5(x+h)−x2+5xh=2xh+h2−5hh=2x+h−5→2x−5,\frac{(x + h)^2 - 5(x + h) - x^2 + 5x}{h} = \frac{2xh + h^2 - 5h}{h} = 2x + h - 5 \to 2x - 5,

so f′(x)=2x−5f'(x) = 2x - 5 for every xx.

  1. Find f−′(1)f'_-(1) and f+′(1)f'_+(1) for f(x)=∣x−1∣+xf(x) = \lvert x - 1 \rvert + x. Is ff differentiable at 11?
Show the answer

For x>1x > 1, f(x)=2x−1f(x) = 2x - 1, slope 22. For x<1x < 1, f(x)=1f(x) = 1, slope 00. So f+′(1)=2f'_+(1) = 2 and f−′(1)=0f'_-(1) = 0. They differ, so ff is not differentiable at 11. It has a corner there.

  1. Classify the point x=0x = 0 for f(x)=x1/3f(x) = x^{1/3} and for g(x)=x4/3g(x) = x^{4/3}.
Show the answer

For ff the quotient is h−2/3→∞h^{-2/3} \to \infty from both sides, a vertical tangent. For gg the quotient is h1/3→0h^{1/3} \to 0, so g′(0)=0g'(0) = 0 and gg is differentiable there.

How this lesson connects

It also uses functions and domains.

Resources for this lesson