Differentiability
On a graph with a sharp corner you cannot draw one single tangent: the road turns suddenly. At smooth points a unique tangent exists, and the function is differentiable there.
f is differentiable at c if the limit of (f(c + h) − f(c)) / h as h → 0 exists. Every differentiable function is continuous, but a continuous function need not be differentiable (it can have a corner).
Formula: f′(c) = lim (h → 0) [f(c + h) − f(c)] / h
Rule 1: Test a corner — Compare the slope on the left and on the right.
- Easy: f(x) = |x| at 0: left slope −1, right slope 1, so it is not differentiable at 0.
- Medium: f(x) = x² at x = 3: [(3 + h)² − 9] / h = 6 + h → 6, so f′(3) = 6.
- Exam-type: f(x) = |x − 2| is not differentiable at x = 2.
Rule 2: Piecewise functions — Check continuity first, then the two slopes.
- Easy: f(x) = x² for x ≤ 1 and 2x − 1 for x > 1: continuous at 1 (both 1), left slope 2, right slope 2, so differentiable.
- Medium: f(x) = x² for x ≤ 1 and 3x − 2 for x > 1: continuous, but left slope 2 and right slope 3, so not differentiable at 1.
- Exam-type: Differentiable ⇒ continuous, so a function with a jump cannot be differentiable there.
Differentiable implies continuous; continuous does not imply differentiable.
Thinking a continuous graph is always smooth.
Smooth = one tangent; a corner = two different slopes.
Practice questions
- 1Is f(x) = |x| differentiable at x = 0?
- 2Find f′(x) for f(x) = x² using the definition.
- 3For f(x) = x², evaluate [(x + h)² − x²] / h at x = 5 as h → 0.
- 4Is every differentiable function continuous?
- 5Is every continuous function differentiable?
- 6Is f(x) = x² for x ≥ 0 and −x² for x < 0 differentiable at 0?
- 7Is f(x) = 2x + 1 for x ≤ 1 and x² + 2 for x > 1 differentiable at 1?
- 8Is f(x) = x² for x ≤ 2 and 3x − 2 for x > 2 differentiable at x = 2?
Show answers (try them yourself first)
- 1No, the slopes −1 and 1 differ
- 2[(x + h)² − x²] / h = 2x + h → 2x
- 32(5) + h → 10
- 4Yes
- 5No, for example |x| at 0
- 6Yes: both one-sided slopes are 0, so f′(0) = 0
- 7Yes: it is continuous (both parts equal 3 at 1) and both slopes are 2
- 8Continuous (both parts are 4) but the slopes are 4 and 3, so not differentiable
Derivatives of composite functions (chain rule)
A bicycle's gears pass the turning of the pedal to the wheel. If the pedal speeds the middle gear by a factor and that gear speeds the wheel by another factor, the total is the product of the factors.
If y = f(g(x)), then dy/dx = f′(g(x)) · g′(x): derivative of the outer function at the inner function, times the derivative of the inner function. Product rule: (uv)′ = u′v + uv′. Quotient rule: (u/v)′ = (u′v − uv′) / v².
Formula: d/dx f(g(x)) = f′(g(x)) · g′(x)
Rule 1: Use the chain rule — Outer derivative first, then multiply by the inner derivative.
- Easy: d/dx (3x + 2)⁵ = 5(3x + 2)⁴ × 3 = 15(3x + 2)⁴.
- Medium: d/dx sin(x²) = cos(x²) × 2x = 2x cos(x²).
- Exam-type: d/dx √(x² + 1) = x / √(x² + 1).
Rule 2: Product and quotient rules — Name u and v first.
- Easy: d/dx (x² sin x) = 2x sin x + x² cos x.
- Medium: d/dx [x / (x + 1)] = 1 / (x + 1)².
- Exam-type: d/dx [sin x / x] = (x cos x − sin x) / x².
Differentiate from the outside in, and multiply as you go.
Forgetting to multiply by the derivative of the inside.
"Derivative of the outside, times derivative of the inside."
Practice questions
- 1Differentiate (2x + 3)⁴.
- 2Differentiate cos 5x.
- 3Differentiate (x³ + x)⁴.
- 4Differentiate sin²x.
- 5Differentiate √(3x + 1).
- 6Differentiate x² cos x.
- 7Differentiate x / (x² + 1).
- 8Differentiate sin(cos x).
Show answers (try them yourself first)
- 14(2x + 3)³ × 2 = 8(2x + 3)³
- 2−5 sin 5x
- 34(x³ + x)³ (3x² + 1)
- 42 sin x cos x = sin 2x
- 53 / (2√(3x + 1))
- 62x cos x − x² sin x
- 7[(x² + 1) − 2x²] / (x² + 1)² = (1 − x²) / (x² + 1)²
- 8−sin x cos(cos x)