MATH 136: Calculus for the Life Sciences II

Subsection 12.1.1 Review of the Chain Rule for Functions of a Single Variable

Recall the chain rule for composite functions: if \(y = f(x)\) and \(x = g(t)\) are both differentiable functions, then \(y\) is a differentiable function of \(t\text{,}\) and $$ \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}. $$

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 \(y\text{,}\) where \(x=g(t)\) and \(y=h(t)\) are both differentiable functions of \(t\text{.}\) Then \(z\) is a differentiable function of \(t\text{,}\) and $$ \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}. $$

Functions of Three Variables:

Suppose that \(w = f(x,y,z)\) is a differentiable function of \(x\text{,}\) \(y\text{,}\) and \(z\text{,}\) and \(x\text{,}\) \(y\text{,}\) and \(z\) are differentiable functions of \(t\text{.}\) Then \(w\) is a differentiable function of \(t\text{,}\) and $$ \frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} + \frac{\partial w}{\partial z} \frac{dz}{dt}. $$

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 $$ x^3 + y^3 = 6xy, $$ shown below.

It is not possible to isolate \(y\) and write it in terms of \(x\text{.}\) To determine \(\displaystyle \frac{dy}{dx}\text{,}\) we can differentiate both sides of the equation with respect to \(x\text{,}\) keeping in mind that \(y\) is a function of \(x\text{:}\) $$ \frac{d}{dx}(x^3 + y^3) = \frac{d}{dx} (6xy). $$ This gives $$ 3x^2 + 3y^2 \frac{dy}{dx} = 6x \frac{dy}{dx} + 6y. $$ Re-arranging and solving for \(\displaystyle \frac{dy}{dx}\) (left as an exercise for the reader) gives $$ \frac{dy}{dx} = \frac{2y-x^2}{y^2-2x}. $$

Subsection 12.1.4 Implicit Differentiation Done Differently

Another way to find \(\displaystyle \frac{dy}{dx}\) for the folium of Descartes, \(x^3 + y^3 = 6xy\text{,}\) is to define the function \(F(x,y) = x^3 + y^3 - 6xy.\) 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 \(x\text{,}\) we have $$ \frac{\partial F}{\partial x} \frac{dx}{dx} + \frac{\partial F}{\partial y} \frac{dy}{dx} = 0. $$

But \(\displaystyle \frac{dx}{dx} = 1\text{.}\) Provided \(\displaystyle \frac{\partial F}{\partial y} \ne 0\text{,}\) we can isolate \(\displaystyle \frac{dy}{dx}\) and get $$ \frac{dy}{dx} = - \frac{ \frac{\partial F}{\partial x} }{ \frac{\partial F}{\partial y} } = - \frac{F_x}{F_y}. $$

Applying this result to the folium of Descartes, given by \(F(x,y) = x^3 + y^3 - 6xy\text{,}\) we have

so that $$ \frac{dy}{dx} = -\frac{F_x}{F_y} = - \frac{3x^2 - 6y}{3y^2 - 6x} = \frac{2y-3x^2}{y^2-2x}, $$ 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 $$ \frac{\partial F}{\partial x} \frac{dx}{dx} + \frac{\partial F}{\partial y} \frac{dy}{dx} + \frac{\partial F}{\partial z} \frac{\partial z}{\partial x} = 0. $$ But \(\displaystyle \frac{dx}{dx} = 1\) and \(\displaystyle \frac{dy}{dx} = 0\) (this is because both \(y\) and \(x\) are independent variables; \(y\) does not depend on \(x\)). We thus have $$ \frac{\partial F}{\partial x} + \frac{\partial F}{\partial z} \frac{\partial z}{\partial x} = 0. $$ Provided \(\displaystyle \frac{\partial F}{\partial z} \ne 0\text{,}\) we can isolate \(\displaystyle \frac{\partial z}{\partial x}\) to get $$ \frac{\partial z}{\partial x} = - \frac{ \frac{\partial F}{\partial x} }{ \frac{\partial F}{\partial z} } = - \frac{F_x}{F_z}. $$

Similarly (left as an exercise to the reader), $$ \frac{\partial z}{\partial y} = - \frac{ \frac{\partial F}{\partial y} }{ \frac{\partial F}{\partial z} } = - \frac{F_y}{F_z}. $$

In the following video, we work through a specific example.

Subsection 12.1.6 Summary

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:

The following references provide a good start for review and further study: