Free AP Calculus AB Differentiation: Definition and Fundamental Properties

AP Calculus AB · Differentiation: Definition and Fundamental Properties · free lesson

The Derivative and Derivative Notation

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

From a number to a function

In the last topic the derivative of ff at one point aa was a number, f′(a)f'(a), the slope of the tangent line there. Work it out at every xx where the limit exists and you get a new function, the derivative function f′f'. It takes in an xx and gives back the slope of the graph of ff at that xx.

Definition.

The derivative of ff is the function

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

at every xx where this limit exists. Where it exists, ff is differentiable at xx. Finding f′f' is called differentiating ff.

The working is the one of the last topic, with the letter xx in place of a number. Take f(x)=x3−3xf(x) = x^3 - 3x. Multiply out f(x+h)−f(x)f(x + h) - f(x), divide by hh, and let hh go to 0:

f(x+h)−f(x)h=3hx2+3h2x+h3−3hh=3x2+3hx+h2−3f′(x)=lim⁡h→0(3x2+3hx+h2−3)=3x2−3\begin{aligned} \frac{f(x + h) - f(x)}{h} &= \frac{ {3hx^{2} + 3h^{2}x + h^{3} - 3h} }{h} = {3x^{2} + 3hx + h^{2} - 3} \\ f'(x) &= \lim_{h \to 0} ({3x^{2} + 3hx + h^{2} - 3}) = {3x^{2} - 3} \end{aligned}

One limit gives the slope at every point at once. Put in numbers:

xx−2-2−1-1001122
f′(x)f'(x), the slope of ff9900−3-30099

Watch the tangent line slide along the graph of ff (blue). Its slope, plotted as a height above the same xx, draws the graph of f′f' (green).

Where the blue curve rises, the tangent slopes up and the green graph is above the axis. At x=−1x = -1 and x=1x = 1 the tangent is flat and f′f' is 0. Between them the curve falls and f′f' is negative.

Notation

If y=f(x)y = f(x), all of these name the derivative function:

f′(x)=y′=dydx=ddxf(x)\begin{aligned} f'(x) &= y' = \frac{dy}{dx} = \frac{d}{dx} f(x) \end{aligned}

Read f′(x)f'(x) as "ff prime of xx" and dydx\frac{dy}{dx} as "d y d x", or "the derivative of yy with respect to xx". The second notation recalls where the derivative comes from: a change in yy over a change in xx, ΔyΔx\frac{\Delta y}{\Delta x}, as the change in xx goes to 0. The value at one point is written f′(2)f'(2) or dydx∣x=2\left.\frac{dy}{dx}\right|_{x = 2}. With other letters, s(t)s(t) has derivative s′(t)=dsdts'(t) = \frac{ds}{dt}.

Warning.

f′(2)f'(2) is the derivative evaluated at 2: first find f′(x)f'(x), then put in 2. It is not the derivative of the number f(2)f(2). For f(x)=x3−3xf(x) = x^3 - 3x, f(2)=2f(2) = 2 but f′(2)=9f'(2) = {9}: the height and the slope are different numbers.

The tangent line

The line tangent to the graph of ff at x=ax = a passes through (a,f(a))(a, f(a)) with slope f′(a)f'(a). In point-slope form,

y−f(a)=f′(a)(x−a)\begin{aligned} y - f(a) &= f'(a)(x - a) \end{aligned}

For f(x)=x3−3xf(x) = x^3 - 3x at x=2x = 2: f(2)=2f(2) = 2 and f′(2)=9f'(2) = {9}, so the tangent line is y−2=9(x−2)y - {2} = {9}(x - 2), or y=9x−16y = 9x - 16. Two numbers are needed, a point and a slope, and the slope is the number f′(a)f'(a), never the expression f′(x)f'(x). Move the point of tangency with the slider.

Reading the graph of f' from the graph of f

Without a formula you can still sketch f′f': wherever the graph of ff rises, f′f' is positive; wherever it falls, f′f' is negative; wherever its tangent is horizontal (a peak, a valley), f′f' is 0; and the steeper the graph, the farther f′f' is from 0. Below, the graph of g(x)=x22−2xg(x) = \frac{x^2}{2} - 2x is drawn with its derivative.

The derivative is g′(x)=x−2g'(x) = {x - 2}. The parabola falls until its lowest point at x=2x = 2 and rises after it; so g′g' is negative for x<2x < 2, 0 at x=2x = 2, and positive for x>2x > 2. The green line crosses the axis exactly below the bottom of the parabola.

Note.

The usual slips on the AP exam: using f(a)f(a) as the slope of the tangent line; writing the tangent line with f′(x)f'(x) instead of the number f′(a)f'(a); taking the graph of ff for the graph of f′f' (a graph of ff that is above the axis does not mean f′f' is positive); and losing the middle term when (x+h)2(x + h)^{2} is multiplied out.

Worked examples

Worked examples

Each example is one kind of question the AP exam asks about the derivative as a function: finding it from the definition for a polynomial, a reciprocal and a square root, using its notation, and writing tangent lines from a formula and from a table. Every number in the working is found by the engine.

Example.

1. The definition, for a polynomial. Use the definition of the derivative to find f′(x)f'(x) for f(x)=2x2−5x+1f(x) = 2x^2 - 5x + 1. Then find f′(3)f'(3).

Write f(x+h)f(x + h) in full, subtract f(x)f(x), divide by hh, and let hh go to 0. The constant 1 cancels in the subtraction.

f(x+h)−f(x)=2(x+h)2−5(x+h)+1−(2x2−5x+1)=4hx+2h2−5hf(x+h)−f(x)h=4x+2h−5f′(x)=lim⁡h→0(4x+2h−5)=4x−5\begin{aligned} f(x + h) - f(x) &= 2(x + h)^2 - 5(x + h) + 1 - (2x^2 - 5x + 1) = {4hx + 2h^{2} - 5h} \\ \frac{f(x + h) - f(x)}{h} &= 4x + 2h - 5 \\ f'(x) &= \lim_{h \to 0} ({4x + 2h - 5}) = {4x - 5} \end{aligned}

So f′(x)=4x−5f'(x) = {4x - 5}, and f′(3)=7f'(3) = {7}: at x=3x = 3 the graph rises 77 units for each unit to the right.

Check: the difference quotient at x=3x = 3 with h=0.001h = 0.001 is 7.0027.002, very close to 77.

Example.

2. The definition, for a reciprocal. Find dydx\frac{dy}{dx} for y=1x+2y = \frac{1}{x + 2} from the definition.

The two fractions on top are put over one denominator first; then the hh on top cancels with the hh below.

1x+h+2−1x+2=(x+2)−(x+h+2)(x+h+2)(x+2)=−h(x+h+2)(x+2)1h(1x+h+2−1x+2)=−1(x+h+2)(x+2)\begin{aligned} \frac{1}{x + h + 2} - \frac{1}{x + 2} &= \frac{(x + 2) - (x + h + 2)}{(x + h + 2)(x + 2)} = \frac{-h}{(x + h + 2)(x + 2)} \\ \frac{1}{h}\left(\frac{1}{x + h + 2} - \frac{1}{x + 2}\right) &= \frac{-1}{(x + h + 2)(x + 2)} \end{aligned}

Now hh can go to 0: the bottom becomes (x+2)(x+2)(x + 2)(x + 2). So

dydx=−1(x+2)2\begin{aligned} \frac{dy}{dx} &= -\frac{1}{(x + 2)^{2}} \end{aligned}

for every x≠−2x \ne -2. At x=−2x = -2 the function itself has no value, so it has no derivative there either. The top is always negative, so every tangent line slopes down, as the graph shows.

Example.

3. The definition, for a square root. Find f′(x)f'(x) for f(x)=xf(x) = \sqrt{x}, where x>0x > 0.

The top x+h−x\sqrt{x + h} - \sqrt{x} will not multiply out. Multiply top and bottom by x+h+x\sqrt{x + h} + \sqrt{x}: the top becomes x+h−x=hx + h - x = h, which cancels.

x+h−xh=(x+h)−xh(x+h+x)=1x+h+x\begin{aligned} \frac{\sqrt{x + h} - \sqrt{x}}{h} &= \frac{(x + h) - x}{h(\sqrt{x + h} + \sqrt{x})} = \frac{1}{\sqrt{x + h} + \sqrt{x}} \end{aligned}

As hh goes to 0 the bottom becomes x+x\sqrt{x} + \sqrt{x}, so

f′(x)=12x\begin{aligned} f'(x) &= \frac{1}{2\sqrt{x}} \end{aligned}

For example f′(4)=14f'(4) = {\frac{1}{4}}. The formula needs x>0x > 0: at x=0x = 0 the bottom is 0 and the graph of x\sqrt{x} starts with a vertical tangent.

Example.

4. A tangent line from a formula. Find the line tangent to the graph of f(x)=1x+2f(x) = \frac{1}{x + 2} at x=0x = 0.

A tangent line needs a point and a slope. The point is (0,f(0))(0, f(0)); the slope is f′(0)f'(0), from the derivative found in Example 2.

f(0)=10+2=12f′(0)=−1(0+2)2=−14\begin{aligned} f(0) &= \frac{1}{0 + 2} = {\frac{1}{2}} \\ f'(0) &= -\frac{1}{(0 + 2)^2} = {-\frac{1}{4}} \end{aligned}

In point-slope form the tangent line is y−12=−14(x−0)y - {\frac{1}{2}} = {-\frac{1}{4}}(x - 0), that is y=−14x+12y = -\frac{1}{4}x + \frac{1}{2}.

Check: near x=0x = 0 the line and the curve can hardly be told apart, and at x=0x = 0 both have height 12\frac{1}{2}.

Example.

5. A tangent line from a table. The function gg is differentiable, and some of its values and those of its derivative are given. Write an equation for the line tangent to the graph of gg at x=3x = 3, and find where it crosses the xx-axis.

xx113355
g(x)g(x)44−2-266
g′(x)g'(x)−1-15533

From the column x=3x = 3: the point is (3,−2)(3, {-2}) and the slope is g′(3)=5g'(3) = {5}. The tangent line is

y−(−2)=5(x−3)\begin{aligned} y - ({-2}) &= {5}(x - 3) \end{aligned}

It crosses the xx-axis where y=0y = 0: 0−(−2)=5(x−3)0 - ({-2}) = {5}(x - 3), so x=175x = \frac{17}{5}. A common slip is to use the slope g(3)=−2g(3) = {-2} or the point (3,5)(3, {5}): the row g(x)g(x) gives the point, the row g′(x)g'(x) the slope.

Example.

6. Where is the tangent horizontal, or parallel to a line? Let f(x)=x2−6x+2f(x) = x^2 - 6x + 2. Find the point where the tangent line is horizontal, and the point where it is parallel to the line y=4x−1y = 4x - 1.

First the derivative, from the definition: f(x+h)−f(x)h=2x+h−6\frac{f(x + h) - f(x)}{h} = {2x + h - 6}, so f′(x)=2x−6f'(x) = {2x - 6}. A horizontal tangent has slope 0; a tangent parallel to y=4x−1y = 4x - 1 has slope 4.

2x−6=0x=3\begin{aligned} 2x - 6 &= 0 \\ x &= 3 \end{aligned}
2x−6=4x=5\begin{aligned} 2x - 6 &= 4 \\ x &= 5 \end{aligned}

The tangent is horizontal at (3,−7)({3}, {-7}), the lowest point of the parabola, and parallel to y=4x−1y = 4x - 1 at (5,−3)({5}, {-3}).

Practice

Answer AP-style questions on the derivative and derivative notation. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise the derivative and derivative notation with new numbers