Lesson Notes By Weeks and Term v4 - SHS 2

PRINCIPLES OF CALCULUS

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Subject: Additional Mathematics

Class: SHS 2

Term: 2nd Term

Week: 10

Grade code: 2.3.1.LI.3

Strand code: 3

Sub-strand code: 1

Content standard code: 2.3.1.CS.1

Indicator code: 2.3.1.LI.3

Theme: CALCULUS

Subtheme: PRINCIPLES OF CALCULUS

Lesson Video

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Performance objectives

Lesson summary

This lesson introduces a powerful calculus technique called Implicit Differentiation. So far, we have mostly dealt with functions written *explicitly*, where `y` is isolated on one side of the equation (e.g., `y = 3x² + 5`). However, many important relationships in science, engineering, and economics are defined *implicitly*, where `x` and `y` are mixed together, like the equation of a circle `x² + y² = 25`. Implicit differentiation allows us to find the derivative `dy/dx` (the gradient of the tangent) for these complex relations without first having to solve for `y`.

Lesson notes

Part A: Explicit vs. Implicit Relations An Explicit Function is one where the dependent variable (usually `y`) is written solely in terms of the independent variable (usually `x`). It's easy to see the input-output relationship. Examples: `y = x³ - 4x`, `y = sin(x) + 2`, `f(x) = √(x - 1)` An Implicit Relation is one where `x` and `y` are mixed together in the equation. It is often difficult or impossible to solve for `y` directly. Examples: `x² + y² = 25`, `x³ + y³ = 6xy`, `sin(y) + x = y²`

Think of it this way: an explicit instruction is direct ("Wash the dishes"). An implicit instruction is indirect ("The sink is full of dirty dishes"). Implicit differentiation is our tool for working with these indirect mathematical relationships. Part B: The Technique of Implicit Differentiation

The core idea is to treat `y` as a function of `x` (i.e., `y = f(x)`) even though we don't know what `f(x)` is. Because of this, whenever we differentiate a term involving `y`, we must apply the Chain Rule.

The Key Rule: When differentiating with respect to `x`, the derivative of `xⁿ` is `nxⁿ⁻¹`. However, the derivative of `yⁿ` is `nyⁿ⁻¹ * dy/dx`. We multiply by `dy/dx` as the "inside" function's derivative.

Evaluation guide