MAT148H1 Chap.6 Derivatives from First Principles and Rules
Derivatives from First Principles and Rules
The derivative as a limit of secant slopes
Join the graph points above x = a and x = b by a straight line, the secant; its slope (f(b) − f(a))/(b − a) is the average rate of change of f between the two points. Letting b approach a turns the average into an instantaneous rate. The derivative is defined as f′(a) = lim x→a (f(x) − f(a))/(x − a), whenever this limit exists, and f is then called differentiable at a.
Writing x = a + h gives the equivalent form lim h→0 (f(a + h) − f(a))/h. Geometrically, the secant lines pivot toward the tangent line, and f′(a) is the tangent slope.
Computing from first principles means evaluating that limit directly.
Direct substitution always gives 0/0, so the work is algebraic: find a common denominator for reciprocals, expand powers for polynomials, or multiply by a conjugate for square roots, until a factor of h cancels. For f(x) = 1/(x + 3) the cancellation leaves −1/((x + h + 3)(x + 3)), and the limit is −1/(x + 3)².
The algebra must be finished before the limit is taken.
Cancelling h is legal only because h ≠ 0 inside the limit; after cancellation the remaining expression is usually continuous in h, so substituting h = 0 is justified by continuity rather than by guesswork. A clean write-up states that reason in a short phrase on the final line.
Product, quotient and chain rules
Once a few derivatives are known, the combination rules generate the rest.
The product rule gives (fg)′ = f′g + fg′. The quotient rule, valid when g(c) ≠ 0, gives (f/g)′ = (f′g − fg′)/g². The chain rule says: if the inner function g has a derivative at a and the outer function f has one at the point g(a), then the composition f ∘ g has derivative g′(a)·f′(g(a)) at a.
In Leibniz notation, dw/dx = (dw/dy)(dy/dx), each factor evaluated at the matching point.
Each rule has a proof from the definition, and the course expects you to reproduce the main ideas.
The quotient rule proof combines the fractions, inserts a zero, the product of f(a) and g(a) taken away and put back, in the numerator so that separate difference quotients for f and g appear, and then uses the fact that g is continuous at a.
That continuity is not an extra assumption: every differentiable function is continuous, a short argument worth knowing by heart.
When the derivative fails to exist
Three standard failures appear.
A corner or cusp makes the one-sided difference quotients converge to different numbers, as with |x| at 0, where they tend to −1 and 1. A discontinuity breaks differentiability outright, since differentiable implies continuous.
A vertical tangent makes the difference quotient grow without bound, as with the cube root at 0. Piecewise functions are the usual test case: check continuity at the join first, then compute both one-sided limits of the difference quotient and compare them.
For a piecewise function such as x² for x ≤ 1 and 3x − 2 for x > 1, the values agree at 1, so the function is continuous there, but the left difference quotient tends to 2 and the right one to 3. Matching values is necessary for differentiability, not sufficient.
Habits that keep calculations correct
State which rule you are using and verify its hypotheses, especially g(c) ≠ 0 for quotients.
Evaluate inner functions before outer ones in a chain rule computation, writing down u(c) and u′(c) as numbers. When a first principles answer is finished, check it against the rules at one specific point: agreement there catches most algebra slips.
What this chapter covers
- 01
Secant slope and the limit definition
- 02
First principles with reciprocals and roots
- 03
Product, quotient and chain rules
- 04
Quotient rule proof from the definition
- 05
Differentiable implies continuous
- 06
Corners, discontinuities and vertical tangents
Worked example · free
First principles derivative of a square root
- 1Write the quotient: (√(2x + 2h + 1) − √(2x + 1))/h, and multiply top and bottom by the conjugate √(2x + 2h + 1) + √(2x + 1).
- 2Simplify the numerator: (2x + 2h + 1) − (2x + 1) = 2h, so the quotient becomes 2/(√(2x + 2h + 1) + √(2x + 1)) after cancelling h ≠ 0.
- 1Let h → 0 using continuity of the square root: g′(x) = 2/(2√(2x + 1)) = 1/√(2x + 1).
- 1Evaluate: g′(4) = 1/√9 = 1/3.
Key terms
- Secant Line
- A secant line is the straight line joining two distinct graph points, one above x = a and one above x = b. Its slope measures the average rate of change of f between a and b.
- Derivative at a Point
- The derivative at a is the limit of difference quotients (f(x) − f(a))/(x − a) as x approaches a, when that limit exists. It equals the slope of the tangent line at a.
- Difference Quotient
- A difference quotient is the expression (f(a + h) − f(a))/h for h ≠ 0. It is the secant slope over a step of size h, and its limit as h tends to 0 defines f′(a).
- Chain Rule
- The chain rule is the formula for differentiating a composition: (f ∘ g)′(c) = f′(g(c))·g′(c). It needs a derivative for the inner function g at c and for the outer function f at the image point g(c).
- Cusp or Corner
- A cusp or corner is a point where a continuous graph changes direction abruptly, so the left and right difference quotients approach different values. The absolute value function at 0 is the standard example.
Derivatives from First Principles and Rules FAQ
Why is |x| not differentiable at 0?
At 0 the difference quotient |h|/h equals −1 for negative h and 1 for positive h. Those one-sided limits disagree, so the two-sided limit defining the derivative fails, even though the graph is continuous there. The graph shows this as a sharp corner.
Does differentiable imply continuous, and is the converse true?
Yes in one direction: if f′(a) exists, then f(x) − f(a) equals a difference quotient times (x − a), which tends to f′(a)·0 = 0. The converse fails, as corners and vertical tangents show. Use the forward direction freely in proofs.
When should I use first principles instead of the derivative rules?
Use the definition whenever a question asks for it explicitly, at the join of a piecewise function, or when proving a rule. Otherwise, the rules are faster and equally rigorous once proved. Many proof questions mix both approaches in one solution.
Exam move
Pick three fresh functions, one reciprocal, one square root and one polynomial, and differentiate each from the definition, writing every algebra line. Then confirm each answer with the rules at a single numerical point. Rewrite the quotient rule proof from memory, marking where continuity of g is used. Finally, build one piecewise function with a corner and prove it is not differentiable using one-sided limits.
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