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Techniques of differentiation

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By the end, you can

  • Interpret a derivative as local rate of change
  • Use product, quotient and chain rules
  • Differentiate implicit relations without losing y dependence

about 25 minutes for the lesson · practise at your own pace.

The central idea

The derivative is the limit of a difference quotient. The chain rule tracks nested dependence: d[f(g(x))]/dx=f′(g(x))g′(x)d[f(g(x))]/dx=f\prime(g(x))g\prime(x). For implicit equations, y depends on x, so differentiating a y-term introduces dy/dxdy/dx. Limits describe approaching a point even when the value there is separately defined.

Limits and continuity

If f(x) approaches L as x approaches a, write lim⁡x→af(x)=L\lim_{x\to a}f(x)=L. For f(x)=x2−x+2f(x)=x^2-x+2, the limit as x approaches 2 is 4.

A limit can exist where the function is undefined. In radians, sin⁡x/x\sin x/x is undefined at zero but has limit 1 there; its graph has a missing point at (0,1)(0,1). Continuity at a requires a defined value and lim⁡x→af(x)=f(a)\lim_{x\to a}f(x)=f(a). The notes’ classroom continuity assumption applies on the interval under consideration, not at excluded inputs.

Approach from both sides

For the source function f(x)=x2−x+2f(x)=x^2-x+2, the two printed value lists approach x=2x=2 from opposite sides:

x from the leftf(x)x from the rightf(x)
1.02.0000003.08.000000
1.52.7500002.55.750000
1.83.4400002.24.640000
1.93.7100002.14.310000
1.953.8525002.054.152500
1.993.9701002.014.030100
1.9953.9850252.0054.015025
1.9993.9970012.0014.003001

Both lists approach 4. Limits underpin calculus, including tangent gradients. The derivative is itself a function: it maps an input x to the gradient of the original function’s tangent at that input, wherever that derivative exists.

A secant approaches a tangent

For P=(x,f(x))P=(x,f(x)) and Q=(x+h,f(x+h))Q=(x+h,f(x+h)), the secant gradient is [f(x+h)−f(x)]/h=Δy/h[f(x+h)-f(x)]/h=\Delta y/h, for h≠0h\ne0. If its limit exists as h→0h\to0, that limit is the tangent gradient f′(x)=dy/dxf^{\prime}(x)=dy/dx. The source uses δx\delta x instead of h and calls the tangent AB. Continuity alone does not guarantee a derivative; the finite limit must exist.

The source’s page 5 QR code points to an exploration of the limiting process. This preserves the original resource pointer; the external activity is not claimed as reviewed. Appendix Examples A–C on PDF 17–20 are included below with their first-principles working and calculator-limit qualifications.

Read the same limit from a curve and a table

Approach the same limit from both sides

For f(x) = x² − x + 2, move towards x = 2 and compare the curve with the table. Here f(2) = 4 as well as the limit being 4.

x² − x + 2

Distance to the limit point
More precision
Compare the approach
Positionxf(x)
Left1.93.71
At point24
Right2.14.31

At distance 0.1, compare both sides. Limit = 4. f(2) = 4.

Values are rounded to 9 decimal places. Use the table to inspect points that overlap on the graph.

Nearby values illustrate the limit; a finite table alone is not a proof.

Teaching and source notes

The retained source uses f(x) = x² − x + 2 and approaches x = 2 from both sides. Substitution gives f(2) = 4. Polynomial continuity justifies this limit; a small table alone is not its proof.

The approach control never substitutes h = 0. Rounded nearby function values need not equal 4 exactly, even when the display rounds them to 4. The linked original retains the printed guide projections.

For y=x²−x+2, both sides of x=2 approach y=4. Dashed projections identify x=2 and f(x)=4.
Open original diagram (new tab)

Distinguish a missing value from a missing limit

A missing value can still have a limit

Use radians. The curve sin(x)/x approaches 1 from both sides, but no function value is assigned at x = 0. The open circle stays open.

sin(x)/x, x ≠ 0 · radians

Distance to the limit point
More precision
View
Compare the approach
Positionxf(x)
Left-0.10.998334166
At point0Undefined
Right0.10.998334166

At distance 0.1, compare both sides. Limit = 1. f(0) remains undefined.

Values are rounded to 9 decimal places. Use the table to inspect points that overlap on the graph.

Near zero zooms the x-axis to separate the sample points from the hole. In the whole view, overlapping sample markers are hidden; the table keeps their values.

Nearby values illustrate the limit; a finite table alone is not a proof.

Teaching and source notes

The source graph has an open circle at (0, 1). No value is assigned to sin(x)/x at x = 0, and the explorer keeps a gap in the curve there. The two-sided limit is 1 when x is measured in radians.

The approach table supports comparison; it is not a proof. Establishing this standard limit requires an argument such as the squeeze theorem. A rounded value of 1 does not remove the domain hole.

The graph of sin(x)/x on [−3π,3π] has an open circle at (0,1). Its limit is 1 as x approaches 0, despite being undefined at 0. Angles are in radians.
Open original diagram (new tab)

Let a secant approach a tangent

Move a secant towards the tangent

Illustrative example: y = x² with P = (1, 1). Move Q = (1 + h, (1 + h)²) towards P. The source sketch uses a general smooth curve, not this particular formula.

Illustrative curve y = x²

Size of h
More precision
Approach from
Points on the curve
Pointxf(x)
P · fixed11
Q · moving1.11.21

The dashed line is the tangent at P; the segment PQ is the secant.

h = 0.1. Secant gradient = 2.1; tangent gradient = 2.

Values are rounded to 9 decimal places. Use the table to inspect points that overlap on the graph.

Nearby values illustrate the limit; a finite table alone is not a proof.

Teaching and source notes

The source sketch is a general smooth curve with P(x, y), Q(x + h, y + Δy), secant PQ and tangent AB. This numerical example is explicitly y = x² at P = (1, 1), not a recovered formula for the original drawing.

For h ≠ 0, Δy = (1 + h)² − 1 = h(2 + h), so Δy/h = 2 + h. Its limit is 2 from either side. At h = 0 the difference quotient is undefined; the derivative comes from its limit. A secant need not tend to a tangent where the derivative does not exist.

An illustrative smooth curve shows P(x,y), Q(x+h,y+Δy), secant PQ and tangent AB. The difference quotient Δy/h tends to the tangent gradient as h approaches 0 if the derivative exists.
Open original diagram (new tab)

Explore

Use the worked example below, then change one input or assumption. Predict which part of the answer must change before calculating.

Related interactive example

Explore this related example, then return to the prediction task above.

Three layers, three factors

Step through the layers from the inside out. At x = 5 the inner value is 1.

Teal: g(5) = 1. No comparison overlay.

Start: g(5) = 1

The innermost layer: g(5)=1g(5)=1 and g′(5)=−3g'(5)=-3. This supplies the factor −3-3.

  • Running product so far
  • Value after this layer
Ideas to try
Which layer changes the sign? Which changes the size most?
Open the full interactive problem →

Worked example

For y=(3x+1)4y=(3x+1)^4, dy/dx=12(3x+1)3dy/dx=12(3x+1)^3. For x2+y2=25x^2+y^2=25, implicit differentiation gives 2x+2y(dy/dx)=02x+2y(dy/dx)=0, hence dy/dx=−x/ydy/dx=-x/y where y is nonzero.

Treating y as a constant in an implicit equation gives the wrong tangent gradient.

Check your understanding

Differentiate y=ex2y=e^{x^2}.

Mastery learning objectives

At the end of this chapter, I should be able to:

Limits

Find the limit of a given function.

First principles

Find the derivative of a given function using first principles.

Rules of differentiation

Apply the chain rule.

Apply the product rule.

Apply the quotient rule.

Methods of differentiation

Use implicit differentiation.

Use logarithmic differentiation.

Different function types

Differentiate powers of x.

Differentiate trigonometric functions.

Differentiate exponential functions.

Differentiate logarithmic functions.

Differentiate inverse trigonometric functions.

Calculator evaluation

Find the approximate value of a derivative at a given point using a graphing calculator.

Choose a practice question

The eight rows below retain the printed page21 question-type map, including its question references. A dash means a blank source cell. Use the separate lecture and tutorial mastery checkpoints above.

Common question typeLecture examplesDiscussionReview follow-up
Approximate a derivative at a point using G.C.141—
Powers of x4, 5, 62(a)5(c)
Trigonometric7, 82(b)(c)1(iii)
Exponential9(a)(b)2(f)1(i), 6(a)
Logarithmic9(c)(d), 102(d)(e), 7(a)(i)2(b), 3(ii), 4(a), 5(b), 6(b), 7(ii), 12(a)
Inverse trigonometric132(g)(h)(i), 6, 7(a)(ii)1(ii), 2(a), 3(i), 4(b), 5(a), 7(i), 12(b)
Implicit differentiation113, 4, 7(b)8, 9, 12(c), 13
Logarithmic differentiation12510, 11

Source-map qualification: Review5(b), printed in the Logarithmic row, asks for the derivative of 52x5^{2x}, an exponential expression. Its retained solution also gives a logarithmic-differentiation method; the source reference is preserved rather than silently moved. Discussion1’s printed key gives12.017, while the existing analytically checked solution explains why12.016 is the correct five-significant-figure derivative. Review13 prints its requested result in the question and has no separate answer on page26.

Practice questions

Practice section 1 · Questions 1–6

Evaluate finite and infinite limits in two ways · Derive the derivative of a square from first principles · +4 more

Practice section 2 · Questions 7–12

Practise power, chain and product rules together · Choose useful trig identities before differentiating · +4 more

Practice section 3 · Questions 13–18

Practise products, exponentials and a logarithmic square root · Collect the derivative terms in implicit relations · +4 more

Practice section 4 · Questions 19–24

Combine a square root with inverse cosine · Evaluate a derivative at a point with the source calculator workflow · +4 more

Practice section 5 · Questions 25–30

Prove the inverse-tangent derivative by implicit differentiation · A nested derivative, correct to five significant figures · +4 more

Practice section 6 · Questions 31–36

An inverse sine derivative with two branches · Single-fraction derivatives and an implicit slope · +4 more

Practice section 7 · Questions 37–42

Use chain, logarithmic, product and quotient differentiation · Differentiate a product with constant-e exponent and a variable power · +4 more

Practice section 8 · Questions 43–45

Logarithmically differentiate an implicit variable power · Compare direct and logarithmic implicit differentiation · +1 more

More practice

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