Derivative as slope of a tangent
A secant is a straight line through two points of a curve. Its slope is rise ÷ run = [f(x + h) − f(x)]/h.
Now slide the second point towards the first (h → 0). The secant turns into the tangent, the line that just touches the curve at one point. Its slope is the derivative f′(x).
On y = x² at x = 1: slope of secant = 2 + h, so the tangent slope is 2.
Derivative as rate of change
If s(t) is distance after time t, then [s(t + h) − s(t)]/h is the average speed over time h. Letting h → 0 gives the instantaneous speed, ds/dt.
In general, dy/dx tells how many units y changes for each one-unit change in x, right at that moment. Example: the area of a circle A = πr² changes at dA/dr = 2πr. At r = 5 cm, that is 10π cm² for each cm.
Derivative by first principle
The definition (also called the first principle or ab-initio method) is:
f′(x) = lim(h→0) [f(x + h) − f(x)]/h
Steps: (1) write f(x + h); (2) subtract f(x); (3) divide by h and simplify; (4) let h → 0.
Standard results
- d/dx(xⁿ) = n xⁿ⁻¹ (any real n)
- d/dx(constant) = 0
- d/dx(sin x) = cos x, d/dx(cos x) = −sin x, d/dx(tan x) = sec²x
For sin x: [sin(x + h) − sin x]/h = 2cos(x + h/2)·sin(h/2)/h. Since sin(h/2)/(h/2) → 1, the answer is cos x. This uses the limit sin θ/θ → 1 from the previous lesson.
Derivative of sum and difference
(u + v)′ = u′ + v′ and (u − v)′ = u′ − v′. Also (k·u)′ = k·u′ for a constant k.
So a polynomial is differentiated term by term: d/dx(4x³ − 5x + 7) = 12x² − 5.
Derivative of a product (product rule)
(uv)′ = u·v′ + v·u′
Picture a rectangle with sides u and v. When x grows a little, u grows by du and v by dv. The area gains a top strip u·dv and a side strip v·du. The tiny corner du·dv is so small that it vanishes as h → 0.
Warning: (uv)′ is not u′·v′.
Derivative of a quotient (quotient rule)
(u/v)′ = (v·u′ − u·v′)/v², where v ≠ 0.
Order matters because of the minus sign: bottom × derivative of top, minus top × derivative of bottom, all over bottom squared.
Example: d/dx [x/(x + 1)] = [(x + 1)·1 − x·1]/(x + 1)² = 1/(x + 1)².
Try it
Walk for 4 minutes and note your distance each minute. Work out average speed for each minute. Then, in the 3D free play, pick x², set x = 1 and slowly shrink h. Say the slope out loud before you look at the readout.
Key formulas and definitions
- f′(x) = lim(h→0) [f(x + h) − f(x)]/h
- d/dx(xⁿ) = n xⁿ⁻¹; d/dx(c) = 0
- d/dx(sin x) = cos x; d/dx(cos x) = −sin x; d/dx(tan x) = sec²x
- (u ± v)′ = u′ ± v′; (ku)′ = k u′
- (uv)′ = u v′ + v u′
- (u/v)′ = (v u′ − u v′)/v²
Worked examples
1. Find the derivative of f(x) = x² by first principle.
f(x + h) − f(x) = x² + 2xh + h² − x² = 2xh + h². Divide by h: 2x + h. Let h → 0: f′(x) = 2x.
2. Find the derivative of f(x) = 1/x by first principle.
[1/(x + h) − 1/x]/h = [x − (x + h)]/[h·x(x + h)] = −1/[x(x + h)]. Let h → 0: f′(x) = −1/x².
3. Differentiate y = 5x⁴ − 3x² + 2x − 9.
Term by term: 20x³ − 6x + 2 − 0 = 20x³ − 6x + 2.
4. A stone falls s = 5t² metres in t seconds. Find its speed at t = 3 s.
ds/dt = 10t. At t = 3: 30 m/s.
5. Differentiate y = x² sin x.
Product rule with u = x², v = sin x: y′ = x²·cos x + sin x·2x = x² cos x + 2x sin x.
6. Differentiate y = (x + 1)/(x − 1).
u = x + 1, v = x − 1, u′ = v′ = 1. y′ = [(x − 1)·1 − (x + 1)·1]/(x − 1)² = −2/(x − 1)².
7. Differentiate y = sin x / x.
Quotient rule: y′ = (x cos x − sin x)/x².
8. Find the slope of the tangent to y = x³ − 2x at x = 2.
y′ = 3x² − 2. At x = 2: 12 − 2 = 10. The tangent slope is 10.
Common mistakes
- Writing (uv)′ = u′v′. The correct rule has two parts: uv′ + vu′.
- Swapping the order in the quotient rule. It is v·u′ − u·v′ on top, not u·v′ − v·u′.
- Letting h = 0 too early in first principle. Cancel h first, then let h → 0.
- Forgetting that the derivative of a constant is 0, or that d/dx(cos x) has a minus sign.