Chapter 10

Further Calculus and Differential Equations (Pure 3)

Differentiating implicitly and parametrically, integration by substitution and by parts, integrating a partial fraction, and separating the variables of a differential equation.

หลักสูตรแกนกลาง / สสวท.: แคลคูลัส (Calculus)เกินหลักสูตรไทยระดับมัธยม — ม.6 เพิ่มเติมหยุดที่อนุพันธ์และปริพันธ์พื้นฐาน ไม่มีการหาอนุพันธ์โดยปริยาย พาราเมตริก หรือสมการเชิงอนุพันธ์
Common Core (US): —
Cambridge: 9709 Pure Mathematics 3 — Differentiation; Integration; Differential equations

10.1 Implicit and parametric differentiation

The two new kinds

\frac{d}{d x} [f (y)] = f^{'} (y) \frac{d y}{d x} \frac{d y}{d x} = \frac{d y / d t}{d x / d t}

Implicit differentiation is the chain rule applied to $y$ because $y$ depends on $x$ . Parametric differentiation is the chain rule with the parameter cancelling.

The derivatives Pure 3 adds

\frac{d}{d x} tan x = sec^{2} x \frac{d}{d x} sec x = sec x tan x \frac{d}{d x} ln x = \frac{1}{x}

ตัวอย่าง

Find $\frac{d y}{d x}$ for the curve $x^{2} + 3 x y - y^{2} = 9$ .

วิธีทำ

1.Differentiate each term: $2 x + 3 (y + x \frac{d y}{d x}) - 2 y \frac{d y}{d x} = 0$
2.Collect the $\frac{d y}{d x}$ terms: $\frac{d y}{d x} (3 x - 2 y) = - 2 x - 3 y$

ตอบ $\frac{d y}{d x} = \frac{2 x + 3 y}{2 y - 3 x}$

ข้อควรระวัง

The $3 x y$ term needs the PRODUCT rule as well as the chain rule, because it is a product of two things that both change. Differentiating it as $3 y$ is the standard slip, and everything after it is then algebra done carefully on a wrong equation.

📖 2 more sections of this chapter are being checked

The whole chapter is written; we read it back section by section before releasing it, because maths nobody has checked should not be on a public page. What is open is complete — nothing here is abridged.

Read it — now practise it

Generate a worksheet on this topic, or sit a timed mock paper.

Build a worksheet Practice arena