Unit 2 · Topic 2.2 Beta

The Derivative and Its Notation

5 min read · freeNot practiced

The last topic found the instantaneous rate of change of a function as a limit of average rates. That limit is so important that it gets its own name, its own notation and its own picture. This page introduces all three, and then turns the rate at one point into a rate at every point.

The derivative at a point

Let ff be a function and aa an input. The derivative of ff at aa is

f′(a)=limh→0f(a+h)−f(a)h,f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h},

provided this limit exists. Read f′(a)f'(a) as "f prime of a". It is the instantaneous rate of change of ff at aa, nothing more and nothing less.

The same limit can be written with the second input called xx instead of a+ha + h. As h→0h \to 0, x=a+hx = a + h goes to aa, and h=x−ah = x - a:

f′(a)=limx→af(x)−f(a)x−a.f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}.

Both forms are secant slopes between (a,f(a))(a, f(a)) and a nearby point, with the nearby point closing in. Learn to recognize both shapes, because the exam often hands you one of these limits and asks what it is.

The picture: the tangent line

The curve y = f(x) with a point at x = a and a second point at x = a + h. A dashed secant line joins them; the run h and the rise f(a + h) minus f(a) are marked. A solid tangent line touches the curve at x = a. The secant slope is the average rate on [a, a + h]; as h shrinks to 0 it approaches the tangent slope f prime of a.
Figure 1. As h→0h \to 0, the secant line through (a,f(a))(a, f(a)) and (a+h,f(a+h))(a + h, f(a + h)) turns into the tangent line at x=ax = a. LevlPrep original diagram.

Each difference quotient is the slope of a secant line through (a,f(a))(a, f(a)). As h→0h \to 0, those secant lines pivot around the fixed point and settle on one line: the tangent line at x=ax = a. Its slope is the limit of the secant slopes, which is f′(a)f'(a). So

  • f′(a)f'(a) is the slope of the tangent line at (a,f(a))(a, f(a)), and
  • the tangent line passes through (a,f(a))(a, f(a)) with that slope.

Point-slope form gives its equation:

y=f(a)+f′(a)(x−a).y = f(a) + f'(a)(x - a).

Two different numbers go into it: the height f(a)f(a) and the slope f′(a)f'(a). Mixing them up is a common slip. If you zoom in on a smooth graph near the point of tangency, the curve looks more and more like its tangent line. That is why the slope of the tangent line deserves to be called the rate of change of the curve at that point.

Worked example 1. Find f′(3)f'(3) for f(x)=x2f(x) = x^2 and write the tangent line at x=3x = 3.

Difference quotient: (3+h)2−9h\frac{(3 + h)^2 - 9}{h}=9+6h+h2−9h= \frac{9 + 6h + h^2 - 9}{h}=6h+h2h= \frac{6h + h^2}{h}=6= 6+h+ h for h≠0h \ne 0.

Limit: f′(3)f'(3)=limh→0(6+h)= \lim_{h \to 0} (6 + h)=6= 6.

Tangent line: the point is (3,f(3))=(3,9)(3, f(3)) = (3, 9) and the slope is 6, so y=9+6(x−3)y = 9 + 6(x - 3).

The derivative as a function

Nothing stops you from doing the same limit at a general input xx. The result is a new function, the derivative function:

f′(x)=limh→0f(x+h)−f(x)h.f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.

Its input is a point of the graph and its output is the slope there. Where the graph of ff rises, f′(x)f'(x) is positive; where it falls, f′(x)f'(x) is negative; where its tangent line is level, f′(x)=0f'(x) = 0.

Worked example 2. Find f′(x)f'(x) for f(x)=x2f(x) = x^2.

f(x+h)−f(x)h\frac{f(x + h) - f(x)}{h}=x2+2xh+h2−x2h= \frac{x^2 + 2xh + h^2 - x^2}{h}=2xh+h2h= \frac{2xh + h^2}{h}=2x= 2x+h+ h.

So f′(x)f'(x)=limh→0(2x+h)= \lim_{h \to 0} (2x + h)=2x= 2x. Check: f′(3)=6f'(3) = 6, as in the first example. One limit now gives every slope of the parabola at once: −4-4 at x=−2x = -2, 0 at x=0x = 0, 2 at x=1x = 1.

Worked example 3. Find f′(x)f'(x) for f(x)=1xf(x) = \frac{1}{x}.

Combine the fractions in the numerator: 1x+h−1x\frac{1}{x + h} - \frac{1}{x}=x−(x+h)x(x+h)= \frac{x - (x + h)}{x(x + h)}=−hx(x+h)= \frac{-h}{x(x + h)}.

Divide by hh: −1x(x+h)\frac{-1}{x(x + h)}. Let h→0h \to 0: f′(x)=−1x2f'(x) = -\frac{1}{x^2}. The answer is negative for every x≠0x \ne 0, and indeed the graph of 1x\frac{1}{x} falls everywhere it is defined.

Notation

Several notations name the same derivative. You must read all of them.

NotationRead asValue at x=ax = a
f′(x)f'(x)f prime of xf′(a)f'(a)
y′y'y primey′(a)y'(a)
dydx\frac{dy}{dx}dy dx, the derivative of y with respect to xdydx|x=a\left.\frac{dy}{dx}\right|_{x=a}
ddx(x2)\frac{d}{dx}\left( x^2 \right)the derivative of x2x^2evaluate the result at aa

The fraction-like form is Leibniz notation. It reminds you that a derivative is a limit of ΔyΔx\frac{\Delta y}{\Delta x}, a change in output over a change in input. But ΔyΔx\frac{\Delta y}{\Delta x} itself, with real changes, is only an average rate. Keep dydx\frac{dy}{dx} (a derivative) and ΔyΔx\frac{\Delta y}{\Delta x} (a secant slope) apart.

A derivative has the units of a rate. If V(t)V(t) is a volume in liters and tt is in minutes, then V′(t)V'(t) is in liters per minute.

Spotting the definition inside a limit

When a limit has the shape limh→0something(a+h)−something(a)h\lim_{h \to 0} \frac{\text{something}(a + h) - \text{something}(a)}{h}, it is a derivative at a point. For example, limh→0(3+h)2−9h\lim_{h \to 0} \frac{(3 + h)^2 - 9}{h} is f′(3)f'(3) for f(x)=x2f(x) = x^2, because 9=f(3)9 = f(3). It is not "00\frac{0}{0}, undefined": 00\frac{0}{0} is only what you see if you substitute before simplifying. Its value is 6.

The next topic estimates derivatives when you have only a table of values or a graph, with no formula to put in the limit.

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