Lesson Notes By Weeks and Term v4 - SHS 2
PRINCIPLES OF CALCULUS
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PRINCIPLES OF CALCULUS
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Subject: Additional Mathematics
Class: SHS 2
Term: 2nd Term
Week: 9
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
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So far, we have been finding the derivatives of functions that are written *explicitly*, like `y = 3x² + 5x - 2`, where `y` is clearly isolated on one side. However, many important relationships in science, engineering, and economics are defined *implicitly*, where `x` and `y` are mixed together, such as the equation of a circle `x² + y² = 25`. In this lesson, we will learn a powerful technique called implicit differentiation to find the derivative `dy/dx` for these types of relations. This skill allows us to find the gradient of any point on complex curves, which is crucial for solving real-world problems involving rates of change, from the movement of objects to changes in economic models.
A. Explicit vs. Implicit Functions An explicit function is one where the dependent variable (usually `y`) is written solely in terms of the independent variable (usually `x`). It's in the form `y = f(x)`. *Example:* `y = 4x³ - 7`, `y = sin(x) + 2x` We can easily find `dy/dx` using our known rules. An implicit relation is one where `x` and `y` are mixed together, and `y` is not isolated. It's often in the form `f(x, y) = c`. *Example:* `x² + y² = 25` (a circle), `x³y + y³x = 10` It can be difficult or impossible to solve for `y` to make it explicit. We need a new method to find `dy/dx`. B. The Core Idea of Implicit Differentiation
The main idea is to treat `y` as a function of `x` (i.e., `y = y(x)`) and then use the Chain Rule whenever we differentiate a term involving `y`.
The Golden Rule: When differentiating with respect to `x`, if you differentiate a term containing `y`, you must multiply the result by `dy/dx`. `d/dx (x²) = 2x` (Standard power rule) `d/dx (y²) = 2y * (dy/dx)` (Power rule, then chain rule) `d/dx (sin(y)) = cos(y) * (dy/dx)` (Trig rule, then chain rule) `d/dx (xy)` requires the Product Rule: `(d/dx(x)) * y + x * (d/dx(y)) = 1*y + x*(dy/dx) = y + x(dy/dx)` C. Step-by-Step Procedure for Implicit Differentiation
To find `dy/dx` for an implicit relation, follow these steps: Differentiate Both Sides: Differentiate every term on both sides of the equation with respect to `x`. Apply Differentiation Rules: Use the Power, Product, Quotient, and Chain rules as needed. Remember the `dy/dx`: Every time you differentiate a term with `y`, multiply by `dy/dx`. Isolate `dy/dx` Terms: Move all terms containing `dy/dx` to one side of the equation (e.g., the left side). Move Other Terms: Move all other terms to the opposite side (e.g., the right side). Factor Out `dy/dx`: Factor `dy/dx` out as a common factor from the terms on its side. Solve for `dy/dx`: Divide both sides by the expression multiplying `dy/dx`.