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Key ideas at a glance

A composite function is a function inside another, such as (2x + 1)⁴. The chain rule differentiates it from the outside in.
The tangent line at a point (x₁, y₁) with slope m is y − y₁ = m(x − x₁).
The normal line is perpendicular to the tangent, so its slope is −1/m when m ≠ 0.
A horizontal tangent occurs where dy/dx = 0.

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Chain rule and tangent formulas

Chain rule (Leibniz form)dy/dx = (dy/du) × (du/dx) when y depends on u and u depends on x.

Trigonometric examplesd/dx [sin(kx)] = k cos(kx), and d/dx [cos(kx)] = −k sin(kx).

Tangenty − y₁ = m(x − x₁), where m = f′(x₁).

Normaly − y₁ = (−1/m)(x − x₁), where m ≠ 0.

Table 1 – Standard Derivative Formulas and Tangent Equations→Chain rule and tangent formulas

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Tangent, normal and horizontal comparison

Line	Slope	How to get it
Tangent	m = f′(x₁)	Differentiate, then substitute x₁
Normal	−1/m	Take the negative reciprocal of the tangent slope
Horizontal tangent	0	Solve f′(x) = 0 for x, then find y

To find the tangent at a point P(x₁, y₁), differentiate the function, substitute x₁ to get the slope m, and write y − y₁ = m(x − x₁).
The normal is perpendicular to the tangent at P, so its slope is −1/m (when m ≠ 0), and its equation is y − y₁ = (−1/m)(x − x₁).
Special cases if m = 0, the tangent is horizontal with equation y = y₁, and the normal is vertical with equation x = x₁. If the slope is undefined, the tangent is vertical with equation x = x₁, and the normal is horizontal with equation y = y₁.
Worked example for y = (2x − 1)/(x + 2) at x = 1, the point is (1, 1/3). The derivative is 5/(x + 2)², so m = 5/9. The tangent is 5x − 9y − 2 = 0. The normal slope is −9/5, which gives 27x + 15y − 32 = 0. Check: (5/9)(−9/5) = −1.

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One short worked example for each idea

Comparison aspect	Main idea	Keywords
Chain rule.	For y = (2x + 1)⁴ dy/dx = 4(2x + 1)³ × 2 = 8(2x + 1)³. At x = 0 dy/dx = 8
Trigonometric chain rule.	For y = sin(2x) dy/dx = 2 cos(2x). At x = 0 dy/dx = 2
Tangent line.	For y = x² at x = 2: the point is (2 4) and the slope is 2x = 4. So y − 4 = 4(x − 2) which gives y = 4x − 4	pointslopegives
Normal line.	For the same curve at (2 the normal slope is −1/4. So y − 4 = −(1/4)(x − 2) which gives y = −x/4 + 4.5	samecurveslopegives
Horizontal tangent.	For y = x² − 4x dy/dx = 2x − 4 = 0 gives x = 2. The point is (2 −4)	givespoint
Tangent from a rule.	For y = x³ − 3x² + 2x + 2 at x = 2 the point is (2 2) and the slope is 2. So the tangent is y = 2x − 2	pointslope

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Lesson summary

Differentiate the outer function first, then multiply by the derivative of the inner function.
To find a tangent, you need a point and a slope. The slope comes from substituting the x-value into dy/dx.
A normal line uses the negative reciprocal of the tangent slope.
A horizontal tangent comes from setting dy/dx = 0.
Check a tangent line by substituting its point into its equation. The point must satisfy the equation, and the slope must match the derivative at that x-value.
Mixing up the chain rule with the product rule is a common error, so identify the outer and inner functions before you differentiate.

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Lesson 2 Summary: Rules of Differentiation, Chain Rule and Tangent Lines