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Lessons for Multivariable Calculus

Mark Fitch

Section 2.5 Chain Rule

Subsection 2.5.1 Chain Rule

On surfaces and in higher dimensions the derivative of a function \(f:\R^n \to \R\) is dependent on the direction as shown in Section 2.4. In that section we selected the direction aribitrarily. The direction can be determined by a path in the surface (or other function). Here we generalize this idea to provide a means for calculating derivatives involving composition of surface functions with curve functions.

Activity 38.

In this activity we calculate directional derivatives where the direction comes from a curve in the surface.
Guido is romping on the surface \(f(x,y)=x^2+xy+y^2.\) He is following the path \(p(t)=(t^2,\sin t).\)
(a)
Calculate\(\nabla f(x,y).\)
(b)
Calculate \(p^\prime(t).\)
(c)
Consider time \(t=0\text{.}\)
(i)
Where is Guido?
(ii)
What is his direction (2D)?
(iii)
What is his instantaneous change of altitude \(f^\prime(x,y)\text{?}\)
(d)
Consider time \(t=\pi/2\text{.}\)
(i)
Where is Guido?
(ii)
What is his direction (2D)?
(iii)
What is his instantaneous change of altitude \(f^\prime(x,y)\text{?}\)
(e)
Consider time \(t\text{.}\)
(i)
Write an expression for Guido’s location at time \(t\text{.}\)
(ii)
Write an expression for Guido’s direction (2D) at time \(t\text{.}\)
(iii)
Write an expression for his instantaneous change of altitude \(f^\prime(x,y)\) at time \(t\text{.}\)
(f)
Find the formula below in your calculations from this activity.
Figure 2.5.1. Chain Rule as Tangent
Note this could also be written \(\nabla f(x(t),y(t)) \cdot (x^\prime(t),y^\prime(t)) \text{.}\) This can be extended to higher dimensional cases.

Checkpoint 2.5.4.

Why would this be called the ‘chain rule’?

Checkpoint 2.5.5.

Calculate \(\frac{\partial f}{dt}\) for \(f(x,y)=\sin x+ \sin y\text{,}\) \(x(t)=\cos t\) with \(y(t)=\sin t\text{.}\)

Checkpoint 2.5.6. Hot and Cold.

Let \(T(x,y)=xy-2\) represent the temperature at the point \((x,y).\) Suppose Guido is walking around an elliptical path given by \(\frac{x^2}{8}+\frac{y^2}{2}=1.\)
(a)
Find a parametric representation of the ellipse. Call this vector valued function \(p(t)\text{.}\)
(b)
Calculate \(\frac{\partial T(p(t))}{dt}.\)
(c)
Use this to find the lowest and highest temperature Guido experiences on this walk.

Subsection 2.5.2 Implicit Differentiation

Activity 39.

(a)
Review implicit differentiation from single variable calculus as needed.
(b)
Differentiate \(y^2=x^2+2y\) with respect to \(x\) (remember \(y\) is implicitly a function of \(x\)).
(c)
Using the chain rule we can show the following. If \(F(x,y)=0\) and the partials of \(F\) are continuous around \((x,y)\) then \(y\) is a function of \(x\) and
\begin{equation*} \frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y}. \end{equation*}
(d)
Use the theorem above to calculate \(\frac{dy}{dx}\) for \(1+xe^y+x^2y^{2}=0.\)
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