MathLabs

Grade 11

Derivative

The derivative measures how fast a quantity is changing at a single instant, built from the limit of average rates of change over shrinking intervals. It gives the slope of a tangent line, the instantaneous velocity of a moving object, and — through a few algebraic rules — a way to differentiate almost any function built from elementary pieces.

IntuitionFrom average speed to instantaneous speed

A car's speedometer reports how fast it is going right now, not its average speed over the whole trip. If a quantity is described by a function y=f(x)y=f(x), the average rate of change between x0x_0 and x0+hx_0+h is the slope of the straight line (the secant line) joining the two points (x0,f(x0))(x_0, f(x_0)) and (x0+h,f(x0+h))(x_0+h, f(x_0+h)): f(x0+h)−f(x0)h\dfrac{f(x_0+h)-f(x_0)}{h}. This ratio depends on how large the step hh is. To capture the rate of change at the single instant x0x_0, shrink the step: let h→0h \to 0 and watch the secant line pivot around (x0,f(x0))(x_0, f(x_0)) until it settles into the tangent line — the best straight-line approximation to the curve at that point.

Interactive plot showing a curve, the secant line through two nearby points at x0 and x0+h, and sliders for x0 and h; as h shrinks toward 0 the secant line rotates to become the tangent line at x0.
The secant line through (x0,f(x0))(x_0, f(x_0)) and (x0+h,f(x0+h))(x_0+h, f(x_0+h)) for a shrinking step hh. Drag hh toward 00 to watch the secant pivot into the tangent line at x0x_0; its slope is f(x0+h)−f(x0)h→f′(x0)\dfrac{f(x_0+h)-f(x_0)}{h} \to f'(x_0).

SchoolThe limit definition of the derivative

Definition: Derivative at a point

Let ff be a function defined on an open interval containing x0x_0. The **derivative of ff at x0x_0**, written f′(x0)f'(x_0), is the limit of the difference quotient as the step shrinks to zero: f′(x0)=lim⁡h→0f(x0+h)−f(x0)hf'(x_0) = \lim_{h \to 0} \dfrac{f(x_0+h)-f(x_0)}{h}, equivalently f′(x0)=lim⁡x→x0f(x)−f(x0)x−x0f'(x_0) = \lim_{x \to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}. When this limit exists (and is finite), ff is called differentiable at x0x_0. Other common notations for the derivative include y′y', dydx\dfrac{dy}{dx}, and dfdx(x0)\dfrac{df}{dx}(x_0).

f′(x0)=lim⁡h→0f(x0+h)−f(x0)h=lim⁡x→x0f(x)−f(x0)x−x0f'(x_0) = \lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h} = \lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}

Geometric meaning. Since the tangent line at x0x_0 is the limit of secant lines, its slope is exactly f′(x0)f'(x_0). The tangent line therefore has equation y=f(x0)+f′(x0)(x−x0)y = f(x_0) + f'(x_0)(x-x_0) — the best linear approximation to the graph of ff near x0x_0. A horizontal tangent (f′(x0)=0f'(x_0)=0) marks a point where the curve is momentarily flat, such as at a local extremum.

y=f(x0)+f′(x0)(x−x0)y = f(x_0) + f'(x_0)(x - x_0)

Physical meaning. If s(t)s(t) is the position of an object at time tt, the average rate of change s(t0+h)−s(t0)h\dfrac{s(t_0+h)-s(t_0)}{h} is average velocity over [t0,t0+h][t_0, t_0+h], and its limit v(t0)=s′(t0)v(t_0)=s'(t_0) is the instantaneous velocity at t0t_0. Differentiating again gives the acceleration a(t)=v′(t)=s′′(t)a(t) = v'(t) = s''(t), the second derivative of position — the rate at which velocity itself is changing.

If ff is differentiable at x0x_0, then ff is continuous at x0x_0.

Why is it true?

A curve with a well-defined tangent slope at x0x_0 cannot jump there: write f(x)−f(x0)=f(x)−f(x0)x−x0⋅(x−x0)f(x)-f(x_0) = \dfrac{f(x)-f(x_0)}{x-x_0} \cdot (x-x_0); as x→x0x \to x_0 the first factor tends to the finite number f′(x0)f'(x_0) while the second tends to 00, so the product tends to 00 and f(x)→f(x0)f(x) \to f(x_0).

Proof

This is exactly the argument above: the limit of a product is the product of the limits when both exist, and lim⁡x→x0f(x)−f(x0)x−x0=f′(x0)\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=f'(x_0) is finite by hypothesis while lim⁡x→x0(x−x0)=0\lim_{x\to x_0}(x-x_0)=0.

The converse is false: continuity does not imply differentiability. The standard counterexample is f(x)=∣x∣f(x) = |x| at x0=0x_0 = 0. It is continuous there, but the difference quotient f(0+h)−f(0)h=∣h∣h\dfrac{f(0+h)-f(0)}{h} = \dfrac{|h|}{h} equals 11 for h>0h>0 and −1-1 for h<0h<0, so the left- and right-hand limits disagree and f′(0)f'(0) does not exist — the graph has a sharp corner, with no single well-defined tangent slope.

UndergraduateDifferentiation rules

Basic differentiation rules
RuleFormulaCondition / note
Constant multiple(cf)′=cf′(cf)' = c f'cc is a constant
Sum / difference(f±g)′=f′±g′(f \pm g)' = f' \pm g'f,gf, g differentiable
Power rule(xn)′=nxn−1(x^n)' = n x^{n-1}n∈Rn \in \mathbb{R}, xx in the domain
Product rule(fg)′=f′g+fg′(fg)' = f'g + fg'f,gf, g differentiable at xx
Quotient rule(fg)′=f′g−fg′g2\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}g(x)≠0g(x) \neq 0
Chain rule(f∘g)′(x)=f′(g(x)) g′(x)(f \circ g)'(x) = f'(g(x))\, g'(x)gg differentiable at xx, ff differentiable at g(x)g(x)
Theorem: Product rule

If f,gf, g are differentiable at xx, then fgfg is differentiable at xx and (fg)′(x)=f′(x)g(x)+f(x)g′(x)(fg)'(x) = f'(x)g(x) + f(x)g'(x).

Why is it true?

Picture f(x)g(x)f(x)g(x) as the area of a rectangle with width f(x)f(x) and height g(x)g(x). As xx moves, both side lengths change at once, so the rectangle's area changes for two reasons simultaneously: the width grows at rate f′(x)f'(x) while the height stays at g(x)g(x), contributing f′(x)g(x)f'(x)g(x); and the height grows at rate g′(x)g'(x) while the width stays at f(x)f(x), contributing f(x)g′(x)f(x)g'(x). A common mistake is to guess (fg)′=f′g′(fg)' = f'g', but that ignores these two separate contributions entirely.

Proof

Write f(x+h)g(x+h)−f(x)g(x)h\dfrac{f(x+h)g(x+h)-f(x)g(x)}{h} and insert −f(x+h)g(x)+f(x+h)g(x)-f(x+h)g(x)+f(x+h)g(x) in the numerator to split it into f(x+h)⋅g(x+h)−g(x)h+f(x+h)−f(x)h⋅g(x)f(x+h)\cdot\dfrac{g(x+h)-g(x)}{h} + \dfrac{f(x+h)-f(x)}{h}\cdot g(x). As h→0h \to 0, the first difference quotient tends to g′(x)g'(x) and f(x+h)→f(x)f(x+h) \to f(x) by continuity, while the second tends to f′(x)f'(x); the sum tends to f(x)g′(x)+f′(x)g(x)f(x)g'(x) + f'(x)g(x).

Theorem: Chain rule

If gg is differentiable at xx and ff is differentiable at g(x)g(x), then f∘gf \circ g is differentiable at xx and (f∘g)′(x)=f′(g(x)) g′(x)(f \circ g)'(x) = f'(g(x))\, g'(x).

Why is it true?

Rates of change compose by multiplying: if y=f(u)y=f(u) changes twice as fast as uu (that is f′=2f'=2), and u=g(x)u=g(x) changes three times as fast as xx (that is g′=3g'=3), then over a small step in xx, uu moves 33 times as far, and yy then moves 22 times as far as uu moved — so yy moves 2×3=62 \times 3 = 6 times as far as xx did. A very common mistake is to forget the inner derivative g′(x)g'(x): ddxsin⁡(x2)≠cos⁡(x2)\dfrac{d}{dx}\sin(x^2) \neq \cos(x^2); the correct derivative is 2xcos⁡(x2)2x\cos(x^2).

Proof

Since ff is differentiable at g(x)g(x), write f(g(x)+k)−f(g(x))=[f′(g(x))+ε(k)] kf(g(x)+k) - f(g(x)) = [f'(g(x)) + \varepsilon(k)]\,k for kk near 00, where ε(k)→0\varepsilon(k) \to 0 as k→0k \to 0. Substituting k=g(x+h)−g(x)k = g(x+h)-g(x), which tends to 00 as h→0h \to 0 by continuity of gg, and dividing by hh gives f(g(x+h))−f(g(x))h=[f′(g(x))+ε(k)]⋅g(x+h)−g(x)h→f′(g(x)) g′(x)\dfrac{f(g(x+h))-f(g(x))}{h} = [f'(g(x))+\varepsilon(k)]\cdot\dfrac{g(x+h)-g(x)}{h} \to f'(g(x))\,g'(x).

The cubic curve y = x cubed minus 3x, rising from bottom left to a local maximum near x = -1, falling to a local minimum near x = 1, then rising again to the top right; an inflection point is marked at the origin.
y=f(x)=x3−3xy = f(x) = x^3 - 3x, with f′(x)=3x2−3f'(x) = 3x^2 - 3. The marked local maximum and minimum sit exactly where f′(x)=0f'(x)=0, i.e. x=±1x=\pm 1: the curve rises while f′(x)>0f'(x)>0 (outside [−1,1][-1,1]) and falls while f′(x)<0f'(x)<0 (inside [−1,1][-1,1]). The marked inflection point at x=0x=0 is where f′′(x)=6xf''(x)=6x changes sign, i.e. where the curve switches from concave down to concave up.

Example: Differentiating a product

Find f′(x)f'(x) for f(x)=x2sin⁡xf(x) = x^2 \sin x.

Solution

Apply the product rule with u(x)=x2u(x) = x^2 and v(x)=sin⁡xv(x) = \sin x: u′(x)=2xu'(x) = 2x and v′(x)=cos⁡xv'(x) = \cos x, so f′(x)=u′(x)v(x)+u(x)v′(x)=2xsin⁡x+x2cos⁡xf'(x) = u'(x)v(x) + u(x)v'(x) = 2x\sin x + x^2\cos x.

Example

A ball is thrown upward from ground level. Its height in metres is h(t)=20t−5t2h(t) = 20t - 5t^2. (a) Find the instantaneous velocity v(t)=h′(t)v(t) = h'(t). (b) When does the ball reach its maximum height? (c) What is the speed at t=1t = 1 s?

Solution

**Step 1 — Differentiate h(t)=20t−5t2h(t) = 20t - 5t^2 using the power rule.** Each term differentiates separately: (20t)′=20(20t)' = 20 and (5t2)′=10t(5t^2)' = 10t, giving v(t)=h′(t)=20−10tv(t) = h'(t) = 20 - 10t.

Step 2 — Find the time of maximum height. At the peak the ball is momentarily at rest, so v(t)=0v(t) = 0: 20−10t=0⇒t=220 - 10t = 0 \Rightarrow t = 2 s. The second derivative h′′(t)=−10<0h''(t) = -10 < 0 confirms this is a maximum.

**Step 3 — Evaluate the speed at t=1t = 1 s.** Substituting into the velocity formula: v(1)=20−10(1)=10v(1) = 20 - 10(1) = 10 m/s upward. The ball is still climbing at t=1t = 1 s, consistent with reaching the peak at t=2t = 2 s.

AdvancedHigher-order derivatives and where this leads

Everything above rests on the limit of a function, so a solid grasp of limits is the prerequisite for differentiation. Differentiating f′f' itself gives the second derivative f′′=(f′)′f''=(f')' (acceleration, curvature), and repeating the process gives f(n)f^{(n)}, the nn-th derivative; a function with continuous derivatives up to order kk is called CkC^k, and C∞C^\infty if it has derivatives of every order. Two directions open up from here: reading off a function's shape — where it increases, has extrema, or is concave — directly from the sign of f′f' and f′′f'' (see applications of the derivative), and running the process in reverse: given f′f', recovering ff is the problem of antiderivatives, the gateway to integral calculus.

ResearchDerivatives at the research frontier

Using the limit definition, what is f′(3)f'(3) for f(x)=x2f(x) = x^2?

Why is f(x)=∣x∣f(x) = |x| continuous but not differentiable at x=0x=0?

By the chain rule, what is the derivative of y=sin⁡(2x2)y = \sin(2x^2)?

A particle moves along a line with position s(t)=t3−6t2+9ts(t) = t^3 - 6t^2 + 9t (meters), t≥0t \ge 0 in seconds. At what time is its acceleration equal to zero?

References

  1. Michael Spivak (2008). Calculus
  2. James Stewart (2015). Calculus: Early Transcendentals
  3. Augustin-Louis Cauchy (1823). Résumé des leçons données à l'École royale polytechnique sur le calcul infinitésimal
  4. Atilim Gunes Baydin, Barak A. Pearlmutter, Alexey Andreyevich Radul, Jeffrey Mark Siskind (2018). Automatic Differentiation in Machine Learning: a Survey