- Compute a derivative from the definition and explain what it means
- Use the table of derivatives and the rules of differentiation
- Differentiate composite and implicit functions
- Find higher derivatives and compute velocity and acceleration
A car's speedometer, the steepness of a mountain road at one spot, the rate at which a bacterial colony grows at noon: all three are the same mathematical object, the instantaneous rate of change. In the previous lesson limits turned 0/0 into a meaningful number; now we use them to define the most-used tool of calculus, the derivative.
Definition and meaning of the derivative
Let y = f(x) and move x₀ by a small step h. The function changes by Δy = f(x₀ + h) − f(x₀), and the average rate of change Δy/h is the slope of the secant through the points (x₀, f(x₀)) and (x₀ + h, f(x₀ + h)). As h → 0 the second point slides towards the first, and the secant turns into the tangent.
- f′(x₀)the derivative at x₀, the slope of the tangent
- h = Δxthe increment of the argument
- f(x₀ + h) − f(x₀) = Δythe increment of the function
If this limit exists, f is differentiable at x₀. Notations: f′(x), y′, dy/dx (Leibniz), and ẋ for derivatives with respect to time (Newton).
Using the definition, find the derivatives of f(x) = x² and g(x) = 1/x.
Show solutionHide solution
g: (1/(x + h) − 1/x)/h = (x − (x + h)) / (x(x + h)h) = −1/(x(x + h)) → −1/x² as h → 0.
So (x²)′ = 2x and (1/x)′ = −1/x², in agreement with the power rule xⁿ → n · xⁿ⁻¹ for n = 2 and n = −1.
Geometric meaning: f′(x₀) is the slope of the tangent, f′(x₀) = tan α, where α is the angle between the tangent and the x-axis. Physical meaning: if s(t) is the position of a body, v(t) = s′(t) is its velocity and a(t) = v′(t) = s″(t) its acceleration. In economics the derivative of cost is the marginal cost, roughly the cost of producing one more unit.
Table of derivatives
| f(x) | f′(x) | Note |
|---|---|---|
| c | 0 | a constant |
| xⁿ | n · xⁿ⁻¹ | any real n: (x⁵)′ = 5x⁴ |
| √x | 1 / (2√x) | x > 0; the case n = 1/2 |
| eˣ | eˣ | equal to its own derivative |
| aˣ | aˣ · ln a | a > 0, a ≠ 1 |
| ln x | 1 / x | x > 0 |
| logₐ x | 1 / (x · ln a) | a > 0, a ≠ 1, x > 0 |
| sin x | cos x | x in radians |
| cos x | −sin x | mind the minus sign |
| tan x | 1 / cos² x | cos x ≠ 0 |
| cot x | −1 / sin² x | sin x ≠ 0 |
| arcsin x | 1 / √(1 − x²) | |x| < 1 |
| arccos x | −1 / √(1 − x²) | |x| < 1 |
| arctan x | 1 / (1 + x²) | all x |
Two rows of the table follow from the remarkable limits: (sin x)′ = lim (sin(x + h) − sin x)/h = lim 2 cos(x + h/2) sin(h/2) / h = lim cos(x + h/2) · sin(h/2)/(h/2) = cos x · 1. And (eˣ)′ = lim (eˣ⁺ʰ − eˣ)/h = eˣ · lim (eʰ − 1)/h = eˣ · 1. This is exactly why e is the “natural” base: with any other base an extra factor ln a appears.
Rules of differentiation
- u, vdifferentiable functions of x
- ca constant
Linearity: the derivative of a sum is the sum of the derivatives, and a constant factor comes out.
- u′, v′the derivatives of the factors
The product rule (Leibniz rule)
- vthe denominator, v ≠ 0
The quotient rule
- gthe inner function, u = g(x)
- fthe outer function, y = f(u)
The chain rule (derivative of a composite function)
Differentiate: a) y = x³ sin x; b) y = (x² + 1)/(x − 1); c) y = sin(3x² + 1); d) y = (2x + 1)⁵.
Show solutionHide solution
b) Quotient rule: y′ = (2x(x − 1) − (x² + 1) · 1)/(x − 1)² = (x² − 2x − 1)/(x − 1)².
c) Chain rule: the outer function is sin u, the inner u = 3x² + 1: y′ = cos(3x² + 1) · 6x = 6x cos(3x² + 1).
d) Outer u⁵, inner 2x + 1: y′ = 5(2x + 1)⁴ · 2 = 10(2x + 1)⁴.
Differentiate: a) ln(x² + 4); b) √(1 + x³); c) e^(sin x); d) arctan 2x.
Show solutionHide solution
a) (1/(x² + 4)) · 2x = 2x/(x² + 4).
b) (1/(2√(1 + x³))) · 3x² = 3x²/(2√(1 + x³)).
c) e^(sin x) · cos x = cos x · e^(sin x).
d) (1/(1 + (2x)²)) · 2 = 2/(1 + 4x²).
Implicit differentiation and higher derivatives
Sometimes y is not given as y = f(x) but by an equation such as x² + y² = 25. Then we differentiate both sides with respect to x, treating y as a function of x (so (y²)′ = 2y · y′ by the chain rule), and solve for y′. The same idea gives logarithmic differentiation: take ln of both sides first, which turns powers and products into sums.
a) Find the slope of the tangent to the circle x² + y² = 25 at the point (3, 4).
b) Differentiate y = xˣ (x > 0).
Show solutionHide solution
At (3, 4): y′ = −3/4. Check: the radius to (3, 4) has slope 4/3 and the tangent is perpendicular to it: (4/3) · (−3/4) = −1. ✓
b) ln y = x ln x. Differentiate: y′/y = ln x + x · (1/x) = ln x + 1.
y′ = xˣ(ln x + 1).
Implicit differentiation also explains the inverse-function rows of the table. Let y = arctan x, so tan y = x. Differentiate: (1/cos² y) · y′ = 1, hence y′ = cos² y = 1/(1 + tan² y) = 1/(1 + x²). In the same way y = ln x gives eʸ = x, then eʸ · y′ = 1 and y′ = 1/eʸ = 1/x.
- f⁻¹the inverse function of f
- f′(x₀)the derivative at x₀, f′(x₀) ≠ 0
Derivative of an inverse function: at corresponding points the slopes of two mutually inverse graphs are reciprocals.
- f″the second derivative, the rate of change of the slope (acceleration in physics)
- f⁽ⁿ⁾the n-th derivative
Useful patterns: (eᵏˣ)⁽ⁿ⁾ = kⁿeᵏˣ; the derivatives of sin x repeat every 4 steps: sin → cos → −sin → −cos → sin.
A body moves along a line according to s(t) = t³ − 6t² + 9t (s in metres, t in seconds). Find v(t) and a(t), the moments when the body stops, and its acceleration at t = 3 s.
Show solutionHide solution
The body stops when v = 0: at t = 1 s and t = 3 s.
a(t) = v′(t) = s″(t) = 6t − 12 m/s².
a(3) = 18 − 12 = 6 m/s². (At t = 1 s, a = −6 m/s²: the body is braking and turns back.)
Key points
- f′(x₀) = lim (h→0) (f(x₀ + h) − f(x₀))/h: the slope of the tangent and the instantaneous rate of change.
- Key derivatives: (xⁿ)′ = nxⁿ⁻¹, (eˣ)′ = eˣ, (ln x)′ = 1/x, (sin x)′ = cos x, (cos x)′ = −sin x.
- Product: (uv)′ = u′v + uv′; quotient: (u/v)′ = (u′v − uv′)/v².
- Chain rule: (f(g(x)))′ = f′(g(x)) · g′(x); never forget the inner derivative.
- For implicit functions differentiate both sides; in motion s′ = v and s″ = a.
Check yourself
10 questions. Every correct answer earns XP.