2026 Fall Section 3.5 The Chain Rule Part II Implicit Diffrentiation

Overview

Professor Ara Araeipour reviews differentiation rules, then develops implicit differentiation as a way to find dy/dx when an equation is not conveniently solved for y. The examples show how the chain rule adds a y′ factor to y-dependent terms, how product-rule expressions are handled, and how the resulting derivative gives tangent-line slopes for curves such as circles and x²y² = 1.

Key takeaways

When differentiating implicitly with respect to x, every differentiated y-dependent expression must include y′; for example, d/dx(y³) = 3y²y′.
Implicit differentiation avoids explicitly solving for y, which can be difficult or produce multiple branches, as with x² + y² = 25.
After applying calculus rules, isolate y′ by moving terms without y′ to the other side and factoring y′ from the remaining terms.
For x² + y² = 25, the derivative is y′ = −x/y, giving slope −3/4 at (3, 4) and tangent line y = −(3/4)x + 25/4.
The product rule still applies when one factor depends on y: d/dx(xeʸ) = eʸ + xeʸy′.
A tangent-line equation requires both a point on the curve and the derivative evaluated at that point; for x²y² = 1 at (−1, 1), the slope is 1 and the line is y = x + 2.

Chapters

0:00 Derivative Rules Needed for Implicit Differentiation

Reviews the derivative definition, power rule, sum and constant-multiple rules, and derivatives of six trigonometric functions.
Recaps product, quotient, and chain rules, including the extended power rule d/dx[g(x)]ⁿ = n[g(x)]ⁿ⁻¹g′(x).
Connects exponential differentiation to the chain rule, including d/dx[eᵍ⁽ˣ⁾] = eᵍ⁽ˣ⁾g′(x).

3:45 Implicit Relations and a Linear Equation Check

Contrasts explicit functions such as y = f(x) with relations such as x² + y² = 25 and x³ + y³ = 6xy.
Defines implicit differentiation as differentiating both sides with respect to x, then solving the resulting equation for y′.
Checks the method on 4x + 2y = 10: explicit solving gives y′ = −2, and implicit differentiation gives 4 + 2y′ = 0, also yielding −2.

7:30 Circle Derivative and Tangent at (3, 4)

For x² + y² = 25, differentiating implicitly gives 2x + 2yy′ = 0, so y′ = −x/y.
At (3, 4), the slope is −3/4; point-slope form gives the tangent line y − 4 = −(3/4)(x − 3), or y = −(3/4)x + 25/4.
Solving explicitly as y = ±√(25 − x²) and differentiating confirms the same derivative on the circle’s upper and lower branches.

16:00 Implicit Differentiation of x³ + y³ = 6xy

Differentiates y³ as 3y²y′ and uses the product rule on 6xy to obtain 6y + 6xy′.
Collecting y′ terms gives 3y²y′ − 6xy′ = 6y − 3x².
Solving yields y′ = (2y − x²)/(y² − 2x), illustrating the combination of chain, product, and algebraic factoring.

19:55 Factoring y′ in Polynomial Implicit Equations

For 4x²y + 2y² equal to a constant, the product rule and chain rule produce 8xy + 4x²y′ + 4yy′ = 0.
Factoring and isolating y′ gives y′ = −2xy/(x² + y).
For 4x² + 6y + y² equal to a constant, differentiation gives 8x + 6y′ + 2yy′ = 0, hence y′ = −4x/(3 + y).

25:40 Product Rule with an Exponential of y

Differentiates the product xeʸ using the product rule; the chain rule gives d/dx(eʸ) = eʸy′.
For xeʸ + 2y³ = 3x, the derivative equation is eʸ + xeʸy′ + 6y²y′ = 3.
Collecting y′ terms gives y′ = (3 − eʸ)/(xeʸ + 6y²).

30:30 Tangent to x²y² = 1 at (−1, 1)

Treats x²y² as a product and differentiates to get 2xy² + 2x²yy′ = 0.
Solving for the derivative gives y′ = −y/x; substituting (−1, 1) gives tangent slope 1.
Using point-slope form, y − 1 = x + 1, so the tangent line is y = x + 2.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Professor Ara Araeipour.