Section 12.1 The Chain Rule and Implicit Differentiation
Subsection 12.1.1 Review of the Chain Rule for Functions of a Single Variable
Recall the chain rule for composite functions: if y=f(x)y=f(x) and x=g(t)x=g(t) are both differentiable functions, then y is a differentiable function of t, and dydt=dydxdxdt.Subsection 12.1.2 The Chain Rule for Functions of Several Variables
Functions of Two Variables: Suppose that z=f(x,y) is a differentiable function of x and y, where x=g(t) and y=h(t) are both differentiable functions of t. Then z is a differentiable function of t, and dzdt=∂z∂xdxdt+∂z∂ydydt. Functions of Three Variables: Suppose that w=f(x,y,z) is a differentiable function of x, y, and z, and x, y, and z are differentiable functions of t. Then w is a differentiable function of t, and dwdt=∂w∂xdxdt+∂w∂ydydt+∂w∂zdzdt. The chain rule extends naturally for functions of more than three variables. In the following video, we work through some examples.Subsection 12.1.3 Review of Implicit Differentiation
Consider the folium of Descartes, the curve defined implicitly by the equation x3+y3=6xy, shown below.
Subsection 12.1.4 Implicit Differentiation Done Differently
Another way to find dydx for the folium of Descartes, x3+y3=6xy, is to define the function F(x,y)=x3+y3−6xy. The folium of Descartes then is given by F(x,y)=0. Using the chain rule from above, and differentiating both sides with respect to x, we have ∂F∂xdxdx+∂F∂ydydx=0. But dxdx=1. Provided ∂F∂y≠0, we can isolate dydx and get dydx=−∂F∂x∂F∂y=−FxFy. Applying this result to the folium of Descartes, given by F(x,y)=x3+y3−6xy, we have
Fx=3x2−6y,Fy=3y2−6x,
so that dydx=−FxFy=−3x2−6y3y2−6x=2y−3x2y2−2x, as before.Subsection 12.1.5 Implicit Differentiation Generalized
Consider a surface z=f(x,y) defined implicitly by an equation of the form F(x,y,z)=0. Using the chain rule from above, and differentiating both sides with respect to x gives ∂F∂xdxdx+∂F∂ydydx+∂F∂z∂z∂x=0. But dxdx=1 and dydx=0 (this is because both y and x are independent variables; y does not depend on x). We thus have ∂F∂x+∂F∂z∂z∂x=0. Provided ∂F∂z≠0, we can isolate ∂z∂x to get ∂z∂x=−∂F∂x∂F∂z=−FxFz. Similarly (left as an exercise to the reader), ∂z∂y=−∂F∂y∂F∂z=−FyFz. In the following video, we work through a specific example.Subsection 12.1.6 Summary
The Chain Rule.
- Suppose that z=f(x,y) is a differentiable function of x and y, where x=g(t) and y=h(t) are both differentiable functions of t. Then z is a differentiable function of t, and dzdt=∂z∂xdxdt+∂z∂ydydt.
- Suppose that w=f(x,y,z) is a differentiable function of x, y, and z, and x, y, and z are differentiable functions of t. Then w is a differentiable function of t, and dwdt=∂w∂xdxdt+∂w∂ydydt+∂w∂zdzdt.
Implicit Differentiation.
- For a curve y=f(x) defined implicitly by the equation F(x,y)=0, dydx=−∂F∂x∂F∂y=−FxFy.
- For a surface z=f(x,y) defined implicitly by the equation F(x,y,z)=0,∂z∂x=−∂F∂x∂F∂z=−FxFz,∂z∂y=−∂F∂y∂F∂z=−FyFz.
Subsection 12.1.7 Don't Forget
Don't forget to return to eClass to complete the pre-class quiz.Subsection 12.1.8 Further Study
Remember that the notes presented above only serve as an introduction to the topic. Further study of the topic will be required. This includes working through the pre-class quizzes, reviewing the lecture notes, and diligently working through the homework problems. As you study, you should reflect on the following learning outcomes, and critically assess where you are on the path to achieving these learning outcomes:Learning Outcomes
- Apply the chain rule for functions of several variables.
- When F(x,y)=0 defines y implicitly in terms of x, determine the derivative of y with respect to x in terms of the partial derivatives of F.
- When F(x,y,z)=0 defines z implicitly in terms of x and y, determine the partial derivative of z with respect to x and the partial derivative of z with respect to y in terms of the partial derivatives of F.
Learning Outcome | Video | Textbook Section |
---|---|---|
1 | 12.E1 | 9.4 |
2 | 12.E2 | 9.4 |
3 | 12.E2 | 9.4 |