Calculus I - Week 5

Implicit Differentiation

Some equations define y as a function of x without solving for y explicitly. This week introduces implicit differentiation — differentiating both sides of an equation with respect to x and solving for dy/dx. It is essential for curves like circles, ellipses, and more complex relations.

Learning Objectives

- Understand when a relation defines y implicitly rather than explicitly. - Differentiate both sides of an equation with respect to x. - Apply the chain rule to terms involving y by writing dy/dx. - Solve the resulting equation for dy/dx. - Find slopes and tangent lines to implicitly defined curves. - Differentiate higher-order implicit equations.

1. Explicit vs. Implicit Functions

An explicit function clearly solves for y: y = x^2 + 1. An implicit function defines a relationship without isolating y. Example: x^2 + y^2 = 25 defines a circle. We cannot write y = sqrt(25 - x^2) for the whole circle — it is not a single function. But we can still find dy/dx using implicit differentiation.

2. The Core Idea

Differentiate both sides of the equation with respect to x. When you differentiate a term involving y, treat y as a function of x. By the chain rule, d/dx[y^2] = 2y * dy/dx. Then collect all dy/dx terms on one side and factor it out.

3. First Example: Circle

Find dy/dx for x^2 + y^2 = 25. Differentiate both sides: d/dx[x^2] + d/dx[y^2] = d/dx[25] 2x + 2y(dy/dx) = 0 Solve for dy/dx: 2y(dy/dx) = -2x dy/dx = -x/y This gives the slope at any point (x, y) on the circle.

4. Second Example: Ellipse

Find dy/dx for 4x^2 + 9y^2 = 36. Differentiate: 8x + 18y(dy/dx) = 0 Solve: dy/dx = -8x / (18y) = -4x / (9y)

5. Tangent Lines to Implicit Curves

Example: Find the tangent line to x^2 + y^2 = 25 at the point (3, 4). From above: dy/dx = -x/y. At (3, 4): slope = -3/4. Tangent line: y - 4 = (-3/4)(x - 3) y = -3x/4 + 9/4 + 4 y = -3x/4 + 25/4.

6. Implicit Differentiation with Products

Example: Find dy/dx for x^2*y + y^3 = 6. Differentiate term by term: d/dx[x^2*y] + d/dx[y^3] = 0 For x^2*y, use the product rule: 2x*y + x^2*(dy/dx) For y^3: 3y^2*(dy/dx) So: 2xy + x^2(dy/dx) + 3y^2(dy/dx) = 0 (x^2 + 3y^2)(dy/dx) = -2xy dy/dx = -2xy / (x^2 + 3y^2).

7. More Complex Example

Find dy/dx for sin(xy) = x + y. Differentiate left side using chain rule: cos(xy) * d/dx[xy] = cos(xy) * (y + x dy/dx) Right side: 1 + dy/dx So: cos(xy)(y + x dy/dx) = 1 + dy/dx y cos(xy) + x cos(xy) dy/dx = 1 + dy/dx Collect dy/dx: x cos(xy) dy/dx - dy/dx = 1 - y cos(xy) (x cos(xy) - 1) dy/dx = 1 - y cos(xy) dy/dx = (1 - y cos(xy)) / (x cos(xy) - 1).

8. Second Derivatives Implicitly

To find d^2y/dx^2, differentiate dy/dx implicitly with respect to x again, substituting back in for dy/dx when needed. Example: For x^2 + y^2 = 25, we have dy/dx = -x/y. Differentiate again: d^2y/dx^2 = d/dx[-x/y] = [-y - (-x)(dy/dx)] / y^2 = (-y + x * (-x/y)) / y^2 = (-y^2 - x^2) / y^3 = -25 / y^3.

9. Application: Related Rates Preview

Implicit differentiation with respect to time t gives related rates. If x^2 + y^2 = 25 and both x and y are functions of time: 2x(dx/dt) + 2y(dy/dt) = 0 If x = 3, y = 4, and dx/dt = 2, then: 2(3)(2) + 2(4)(dy/dt) = 0 12 + 8(dy/dt) = 0 dy/dt = -3/2.

10. Common Mistakes

- Forgetting to write dy/dx when differentiating a y term. - Treating y as a constant instead of a function of x. - Not using the product rule when differentiating xy or x^2*y. - Not factoring out dy/dx before solving. - Plugging in the point too early before solving for dy/dx algebraically.