Differentiation measures how fast a function is changing at one exact moment.
Start from the slope of a secant line through two points, the average rate of change, and shrink the gap between them to 0: it becomes the slope of the tangent line touching the curve at one point, the instantaneous rate of change.
That tangent slope is the derivative at the point, and the function pairing every x with its derivative is the derivative function.
In the first picture, shrinking Δx brings the blue secant onto the orange tangent; in the second, f(x) above and its derivative f'(x) below can be compared at the same x.
The idea of the derivative
💡 What differentiation is
①It finds the rate of change at a single instant
②Letting Δx → 0 in the average rate of change (secant slope) gives the instantaneous rate of change (tangent slope)
③So the slope at a single point on a curve can be found exactly
④Differentiable implies continuous, but not the other way round: y = |x| is continuous at x = 0 but has a corner there and is not differentiable
From secant to tangent
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1.5
seegongsik.com
On f(x) = x², the secant (blue) through a and the point Δx away, and the tangent (orange); as Δx shrinks toward 0, the secant comes onto the tangent
Definition of Derivative
f'(a) = limΔx→0f(a + Δx) - f(a)Δx
The instantaneous rate of change at x = a, which is the slope of the tangent
🔑 How a secant becomes a tangent
①A secant passes through two points on the curve, and its slope is the average rate of change
②For f(x) = x², the secant slope is 2a + Δx
③As Δx → 0, the secant slope approaches 2a, which is the tangent slope f'(a)
Basic differentiation rules
Power rule
(xn)' = nxn-1
Here n is a natural number; real exponents come later in calculus
Constant multiples, sums and differences
(cf)' = cf', (f ± g)' = f' ± g'
Constant multiples come out; sums and differences are differentiated term by term
Product rule
(fg)' = f'g + fg'
Differentiate the first and keep the second, plus keep the first and differentiate the second
Quotient rule
(fg)' = f'g - fg'g²
(derivative of top × bottom − top × derivative of bottom) ÷ bottom squared; covered later in calculus
What the derivative function means
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seegongsik.com
Top: f(x) = x³/3 − x; bottom: its derivative f'(x) = x² − 1. At the same x, the value below is the slope of the curve above
💡 How f(x) and f'(x) are related
①If f'(x) > 0, f(x) is on an increasing interval
②If f'(x) < 0, f(x) is on a decreasing interval
③Points with f'(x) = 0 are candidates for a maximum or minimum; in the picture they are at x = ±1
Summary
Derivative
f'(a) = limh→0f(a+h) - f(a)h
The slope of the tangent at point a
🎯 Exam Points
①Differentiable implies continuous, not the reverse (y = |x| is not differentiable at x = 0)
②(xⁿ)' = nxⁿ⁻¹ is the basis for differentiating polynomials
③Product rule: f'g + fg'
④Whether a point with f'(a) = 0 is an extremum depends on a change of sign
⑤Tangent line: y − f(a) = f'(a)(x − a)
Worked Examples & Past Exam
Example 1
For f(x) = x², use the definition of the derivative to find f'(2).
1
Put a = 2 into the definition of the derivative.
f'(2) = limh→0(2+h)2 - 22h
2
(2 + h)² − 4 = 4h + h², so cancel h and then let h → 0.
= limh→04h + h2h = limh→0 (4 + h) = 4
▸ f'(2) = 4
Before cancelling, the denominator would be 0, so h = 0 cannot be substituted directly. Simplifying to 4 + h and then taking the limit is how a derivative is found from the definition.
Example 2
Find the derivative f'(x) of f(x) = x³ − 2x² + 5.
1
Apply (xⁿ)' = nxⁿ⁻¹ to each term.
(x3)' = 3x2, (2x2)' = 4x, (5)' = 0
2
Add the derivatives of the terms. The derivative of the constant 5 is 0.
f'(x) = 3x2 - 4x
▸ f'(x) = 3x² − 4x
A constant term only shifts the graph up or down and does not change the slope, so it disappears when differentiated.
2023 CSAT Math type, adapted
For f(x) = x² + 3x, what is the derivative f'(1) at x = 1?