The definition of the derivative
The derivative is a limit before it is a rule. Working through the difference quotient by hand is what makes later shortcuts meaningful, and it is where the link between slope, instantaneous rate of change and the tangent line is established. It is also where differentiability is tested: at a corner, a cusp, a vertical tangent or a join where the pieces do not meet, the limit fails to exist as a real number.
In this topic
- Taught
- Before the midterm
- Questions
- 27
- Total marks
- 67
- Suggested time
- 117 min
- Difficulty
Introductory to Capstone
Show the 6 hardest
Key results
- Average rate of change
For ; it is the slope of the secant line through and .
- The derivative at a point
When the limit exists as a real number; then is differentiable at , while a limit of where is continuous, as for at , is a vertical tangent and not a derivative.
- The tangent line
When is differentiable at .
- One-sided slopes
Both one-sided limits must be real numbers; at a corner, such as at 0, they are different.
- Differentiable means continuous
Read the theoryUBC Math 100 §3.1 Review: linesUBC Math 100 §3.2 Slopes and rates of changeUBC Math 100 §3.3 The derivativeUBC Math 100 §3.5 Derivatives of exponential functionsCLP-1 §1.1 Drawing tangents and a first limitCLP-1 §1.2 Another limit and computing velocityCLP-1 §2.1 Revisiting tangent linesCLP-1 §2.2 Definition of the derivativeCLP-1 §2.3 Interpretations of the derivativeCLP-1 §3.1 Velocity and acceleration
Worked example
Use the limit definition of the derivative to find for .
Rules such as the power rule earn no marks on this question.
Strategy
Substitute into the difference quotient, expand carefully, simplify until the h in the denominator cancels, and only then let h approach 0.
Working
Replace every x with 3 + h, including inside the linear term.
Expand and collect like terms.
Evaluate the function at the point itself.
Subtract, then cancel. The h in the denominator is gone, so the limit is now routine.
Let h approach 0. The tangent at x = 3 rises 2 units for every unit to the right.
Answer
The answer is 2.
Check
A small step gives f(3.01) − f(3) = 0.0201, a slope of 2.01 over 0.01, close to 2.
Where this goes wrong
Expanding (3 + h)² correctly but writing −4(3 + h) as −12, or as −12 + 4h. The −4 multiplies both terms of the bracket, giving −12 − 4h; distributing over the whole bracket is where this question is usually lost.
On the paper
Write the definition first and keep on every line until cancels; only then let . Where the definition is asked for, a shortcut rule earns nothing.
Leans on
Expanding binomials, from Functions, graphs and inverses
Your turn
Use the definition of the derivative to find for the function below.
Practice questions
5 of the 27, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.
- Introductory2 marks · 3 min
Which limit represents for ?
Answer this questionStrategy
Substitute the function and the point into the definition: the numerator is g(9 + h) − g(9), with the root over the whole input, and the denominator is the change in the input, h.
Working
Write the definition with a = 9 before touching the square root.
Evaluate both pieces. The second is a constant.
Assemble. The denominator must be h, matching the variable in the limit.
Answer
The answer is the limit as h → 0 of (√(9 + h) − 3)/h.
Check
At h = 0.01 this quotient is about 0.1666, close to the slope 1/6 that the power rule gives for √x at 9.
Where this goes wrong
Choosing an option whose denominator is 9 or 9 + h rather than h, or one that reads g(9 + h) as √9 + h. The quotient divides the change in g by the change in the input, and the root covers the whole input.
Leans on
Limit notation, from Limits
Try a similar question
Use the limit definition of the derivative to find for . - Introductory1 mark · 2 min
A particle moves along a line with position metres at time seconds. Find its average velocity over the interval from to .
Give the answer in metres per second.
Answer this questionStrategy
Average velocity over an interval is the change in position divided by the change in time, which is also the slope of the secant line through the two points on the position graph.
Working
Position at each end of the interval.
Displacement over elapsed time.
Answer
The answer is 9 metres per second.
Check
For a quadratic position the average velocity over an interval is the velocity at its midpoint, and s′(2.5) = 2·2.5 + 4 = 9.
Where this goes wrong
Dividing by 4 instead of by the length of the interval, 3, or averaging the two positions. Average velocity is a difference quotient, not an average of values.
Leans on
Slope of a line through two points, from Functions, graphs and inverses
- Standard3 marks · 5 min
Let . Use the definition of the derivative to find , and then find an equation of the tangent line to at .
Rules such as the chain rule earn no marks for the slope.
Answer this questionStrategy
A square root in the difference quotient calls for the conjugate, which turns the numerator into a multiple of h. With the slope found, the point (2, 2) and point-slope form give the line.
Working
Set up the definition at a = 2.
Conjugate, difference of squares, then cancel h.
Let h → 0.
Point-slope form through (2, 2), then solved for y.
Answer
The answer is y equals negative three quarters x plus seven halves.
Check
The line passes through (2, 2) on the curve, and at x = 2.04 it gives 1.97, against f(2.04) = √3.88 ≈ 1.9698.
Where this goes wrong
Writing the difference of squares as h instead of −3h. The numerator is (4 − 3h) − 4 = −3h; dropping the −3 gives slope 1/4 and a line that rises where the curve falls.
Leans on
Rationalising with a conjugate, from Limits
- Challenging4 marks · 6 min
Find the values of the constants and for which the function below is differentiable everywhere, and enter .
Answer this questionStrategy
Away from x = 1 both pieces are polynomials, so only the join matters. At the join, continuity gives one equation and equal one-sided slopes give the other.
Working
The derivative of 2x² + 1 at 1, from the definition.
The limit exists only if the right piece also equals 3 at x = 1: that is continuity.
Matching slopes gives a; continuity gives b.
Answer
The answer is a is 4 and b is −1.
Check
With a = 4 and b = −1, 2x² + 1 − (4x − 1) = 2(x − 1)², so the line touches the parabola at x = 1 without crossing it.
Where this goes wrong
Matching the slopes but not the values, which gives a = 4 with b unconstrained. A function has to be continuous at a point before it can be differentiable there.
Leans on
Continuity at a join, from Continuity
- Capstone5 marks · 10 min
Let for and . Using the definition of the derivative, show that exists and find its value.
Answer this questionStrategy
Nothing but the definition works here, because the formula has no derivative rule to apply at 0. The difference quotient reduces to h·sin(1/h), a bounded factor times h, and h tends to 0.
Working
Set up the definition at 0; one power of h cancels.
Because |sin(1/h)| ≤ 1 for every h ≠ 0.
A quantity whose size is at most |h| is forced to 0 along with |h|.
The limit that defines the derivative exists and equals 0.
Answer
The answer is the derivative at 0 exists and equals 0.
What earns the marks
- Writes the difference quotient at 0 and simplifies it to h·sin(1/h), noting that f(0) = 0.
- Bounds the quotient: |h·sin(1/h)| ≤ |h|, because |sin(1/h)| ≤ 1.
- Concludes that the quotient tends to 0 because |h| does, so f′(0) exists and equals 0.
Check
The graph is squeezed between y = x² and y = −x², both flat at 0, so its slope at 0 is trapped at 0 too.
Where this goes wrong
Declaring that f′(0) does not exist because sin(1/h) oscillates. The oscillation is real, but it is multiplied by h, which crushes it: its size is at most |h|, and |h| tends to 0.
Leans on
Bounded times vanishing, from Limits
Try a similar question
Evaluate the limit.
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