MCV4U The Derivative Function: Notes and Practice

The Derivative Function

Find the tangent slope at every x at once and you get a new function, the derivative f′(x). Its graph can be built from the slopes of f, its equation comes from first principles with x left as a variable, its degree is one less than f's for a non-constant polynomial, and it fails to exist at corners, cusps, breaks and vertical tangents.

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Key things to remember

1Derivative function:

f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h};

it gives the tangent slope at every

x

2Notation:

f^{'} (x)

y^{'}

\frac{d y}{d x}

and

\frac{d}{d x} f (x)

all mean the derivative.

3For polynomials the degree drops by one: linear gives constant, quadratic gives linear, cubic gives quadratic.

4Where

f

rises,

f^{'} (x) > 0

; where

f

falls,

f^{'} (x) < 0

(apart from single flat points, like

x^{3}

0

); a smooth local maximum or minimum has

f^{'} (x) = 0

5No derivative at a corner, cusp, discontinuity or vertical tangent. Differentiable implies continuous, but not the reverse.

So far you have found the slope of the tangent at one point at a time: $f^{'} (3)$ , then $f^{'} (4)$ , each with its own page of algebra. But a curve has a tangent slope at every point, so the slopes themselves form a function: input an $x$ , output the slope of the tangent there. Do the first-principles algebra once with the letter $x$ in place of a number, and you get a formula for every slope at once. That formula is the derivative function $f^{'} (x)$ .

f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}

The derivative function. The x stays a variable through the algebra; only h goes to 0.

Fig 1Top: f(x) = x² − 4x + 3 with tangents at x = 0, 1, 2, 3 and 4, whose slopes are −4, −2, 0, 2 and 4. Bottom, on the same x-axis: those slopes plotted as points, which lie on the line f′(x) = 2x − 4. At the vertex x = 2 the tangent is horizontal and f′(2) = 0.

Read the figure from left to right. Where the parabola falls, its tangents slope downward and the points on the $f^{'}$ graph are below the $x$ -axis. At the vertex the tangent is flat, so $f^{'}$ crosses zero. Where the parabola rises, the slopes are positive and growing. Plotting slope against $x$ turns the parabola into a straight line.

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Pick a function (such as x², x³ − 3x, sin x or eˣ, or set your own cubic), drag the point along the graph, and switch on Trace f′ to plot each tangent slope as a point. Reveal the full f′ to check your trace.

Finding f′(x) from first principles

f (x + h) - f (x) \frac{f ( x + h ) - f ( x )}{h} = [(x + h)^{2} - 4 (x + h) + 3] - [x^{2} - 4 x + 3] = x^{2} + 2 x h + h^{2} - 4 x - 4 h + 3 - x^{2} + 4 x - 3 = 2 x h + h^{2} - 4 h = 2 x + h - 4 ⟶ 2 x - 4 as h \to 0

For f(x) = x² − 4x + 3: every term without h cancels, the rest is divided by h, and the limit leaves f′(x) = 2x − 4, matching the figure.

Notation: the derivative of $y = f (x)$ can be written $f^{'} (x)$ , $y^{'}$ , $\frac{d y}{d x}$ or $\frac{d}{d x} f (x)$ . The form $\frac{d y}{d x}$ (Leibniz notation) is a reminder that a derivative is a limit of $\frac{Δ y}{Δ x}$ .
To get the slope at one point, find $f^{'} (x)$ once, then substitute: $f^{'} (2) = 2 (2) - 4 = 0$ and $f^{'} (5) = 6$ .

VideoMario applies the difference quotient with x left as a variable to get a whole derivative function, then uses it to find slopes along the graph, the same process as this section.Watch on YouTube

Patterns: the degree drops by one

f(x)	Type of f	f′(x) from first principles	Type of f′
$5$	constant	$0$	zero
$3 x + 2$	linear	$3$	constant
$x^{2} - 4 x + 3$	quadratic	$2 x - 4$	linear
$x^{3} - 6 x$	cubic	$3 x^{2} - 6$	quadratic

For a non-constant polynomial, the derivative is a polynomial of one lower degree; a constant has derivative 0.

Reading the sign of f′ from the graph of f

Where the graph of $f$ rises from left to right, its tangents slope up, so $f^{'} (x) > 0$ and the graph of $f^{'}$ is above the $x$ -axis. (A rising graph can level off for an instant, like $y = x^{3}$ at $x = 0$ , where $f^{'} = 0$ at that single point.)
Where the graph of $f$ falls, $f^{'} (x) < 0$ and the graph of $f^{'}$ is below the $x$ -axis.
At a smooth peak or valley (a local maximum or minimum) the tangent is horizontal, so $f^{'} (x) = 0$ . Between a peak and the next valley the graph falls, so the rate of change is negative (apart from any single flat point); between a valley and the next peak it is positive.
Steeper parts of $f$ give values of $f^{'}$ farther from zero; flatter parts give values near zero.

Where a derivative does not exist

Fig 2Four graphs with no single tangent slope at x = 0: a corner on y = |x| (left slope −1, right slope 1), a cusp on y = x to the power 2/3 (slopes heading to −∞ and +∞), a jump where the graph breaks, and a vertical tangent on the cube-root graph.

A function is differentiable at $x = a$ when $f^{'} (a)$ exists. It fails at a corner and at a cusp (the one-sided slopes disagree), at a break such as a jump (no single tangent can follow a graph that is not continuous) and at a vertical tangent (a vertical line has undefined slope). For $f (x) = ∣ x ∣$ at $0$ the difference quotient is $\frac{∣ h ∣}{h}$ , which is $1$ for every $h > 0$ and $- 1$ for every $h < 0$ , so the limit does not exist. The lesson: a function that is differentiable at a point must be continuous there, but a continuous function, like $∣ x ∣$ , need not be differentiable.

VideoKhan Academy looks at graphs to decide where a function is not differentiable, including discontinuities, sharp corners and vertical tangents.Watch on YouTube

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Use first principles to find

f^{'} (x)

for

f (x) = 3 x^{2} - 5 x + 2

, then find the slope of the tangent at

x = 2

Example 2

Use first principles to find the derivative of the cubic

f (x) = x^{3} - 6 x

Example 3

The graph of

f (x) = 6 x - x^{2}

is a parabola opening down with vertex

(3, 9)

. Describe the graph of

f^{'}

, then confirm with first principles.

Example 4

Show that

f (x) = ∣ x ∣

is continuous but not differentiable at

x = 0

Where marks get lost

Substituting a number for

x

before taking the limit when the question asks for

f^{'} (x)

. Keep

x

as a variable; substitute only at the end, if a slope at a point is wanted.

Reading the HEIGHT of

f

as the height of

f^{'}

. Where

f

is high but flat,

f^{'}

is near

0

; where

f

is below the axis but rising,

f^{'}

is positive.

Thinking a zero of

f

is a zero of

f^{'}

. The zeros of

f^{'}

are where the tangent is horizontal (peaks, valleys and flat spots), not where

f

crosses the

x

-axis.

Assuming every continuous function is differentiable. The graph of

y = ∣ x ∣

is unbroken, but it has no tangent slope at the corner

x = 0

Key things to remember

Worked examples

Where marks get lost

Flashcards 8

Practice