Partial derivatives

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We introduce partial derivatives.

Given a function $F:\R ^n \to \R$ , it is often useful to differentiate with respect to a single variable and hold the other variables as constants. As a concrete example, let $F(x,y) = x^2+2y^2$ Fixing $y=2$ , allows us to focus our attention to all points on the surface where the $y$ -value is $2$ ,

We can now focus our attention on the curve

and differentiate this curve purely with respect to $x$ .

Given a function $F:\R ^n\to \R$ , the partial derivative of $F$ with respect to the $i$ th variable is denoted: $\pp {x_i} F(\vec {x})$ This means that one should take the derivative with respect to $x_i$ while treating all other variables as constants.

Let $F(x,y) = x^2+2y^2$ . Compute: $\pp {x} F(x,y) \begin {prompt} = \answer {2x+0} \end {prompt}$

Compute $\pp {y} F(x,y) \begin {prompt} = \answer {0+4y} \end {prompt}$

There are several different notations for the partial derivative. The two we’ll mainly be using are $\begin{align*} \pp{x} F(x,y,z) &= F^{(1,0,0)}(x,y,z),\\ \pp{y} F(x,y,z) &= F^{(0,1,0)}(x,y,z),\\ \pp{z} F(x,y,z) &= F^{(0,0,1)}(x,y,z). \end{align*}$ Other texts sometimes use $F_x$ to mean $\pp [F]{x}$ , but we won’t use that notation.

Let $F(x,y) = x^2y + 2x+y^3$ . Compute: $F^{(1,0)}(x,y) \begin {prompt} = \answer {2xy+2} \end {prompt}$

Compute: $F^{(0,1)}(x,y) \begin {prompt} = \answer {x^2+3y^2} \end {prompt}$

We have shown how to compute a partial derivative, but it may still not be clear what a partial derivative means. Given $z=F(x,y)$ , $F^{(1,0)}(x,y)$ measures the rate at which $z$ changes as only $x$ varies: $y$ is held constant.

Imagine standing in a rolling meadow, then beginning to walk due east. Depending on your location, you might walk up, sharply down, or perhaps not change elevation at all. This is similar to measuring $\pp [z]{x}$ : you are moving only east (in the “ $x$ ”-direction) and not north/south at all. Going back to your original location, imagine now walking due north (in the “ $y$ ”-direction). Perhaps walking due north does not change your elevation at all. This is analogous to $\pp [z]{y}=0$ : $z$ does not change with respect to $y$ . We can see that $\pp [z]{x}$ and $\pp [z]{y}$ do not have to be the same, or even similar, as it is easy to imagine circumstances where walking east means you walk downhill, though walking north makes you walk uphill. The next example helps us visualize this.

Let $z=F(x,y)=-x^2-\frac {y^2}{2}+xy+10$ . Find $F^{(1,0)}(2,1)$ and $F^{(0,1)}(2,1)$ .

Write with me $\pp {x}F(x,y) = \answer [given]{-2x+y}$ and $\pp {y}F(x,y) = \answer [given]{-y+x}$ Thus $F^{(1,0)}(2,1) = \answer [given]{-3}$ and $F^{(0,1)}(2,1) = \answer [given]{1}$ .

Whenever we do a computation in mathematics, we should ask ourselves, “What does this mean?”

What is the meaning of $\begin{align*} F^{(1,0)}(2,1) &= -3\\ F^{(0,1)}(2,1)&=1? \end{align*}$ First note that $F(2,1) = \answer {7.5}$ . If $F^{(1,0)}(2,1)=-3$ , this means if one “stands” on the surface at the point $(\answer {2},\answer {1},\answer {7.5})$ and moves parallelorthogonal to the $x$ -axis (so only the $x$ -value changes, not the $y$ -value), then the instantaneous rate of change in $z$ is $\answer {-3}$ . Increasing the $x$ -value will increasedecrease the $z$ -value; decreasing the $x$ -value will increasedecrease the $z$ -value.

If $F^{(0,1)}(2,1)=1$ , this means if one “stands” on the surface at the point $(\answer {2},\answer {1},\answer {7.5})$ and moves to the $y$ -axis (so only the $y$ -value changes, not the $x$ -value), then the instantaneous rate of change in $z$ is $\answer {1}$ . Increasing the $y$ -value will the $z$ -value; decreasing the $y$ -value will the $z$ -value.

Finally, since the magnitude of $\pp [F]{x}$ is greater than the magnitude of $\pp [F]{y}$ at $(2,1)$ , the surface is “steeper” in the $x$ -direction than in the $y$ -direction.

Estimating partial derivatives

Functions of several variables, especially ones that map $\R ^2\to \R$ can be described by a table of values or level curves. In either case we can estimate partial derivatives by looking at $\frac {\text {change in the output}}{\text {change in the variable}}$ Let’s do an example to make this more clear.

Let $F:\R ^2\to \R$ be a differentiable function described by the following table of values:

Estimate $F^{(1,0)}(2,6)$

To estimate $F^{(1,0)}(2,6)$ , we examine the change in $F(x,6)$ between $x=1$ and $x=2$ : $\begin{align*} \frac{F(2,6)-F(1,6)}{2-1}&= \frac{\answer[given]{16}-\answer[given]{24}}{2-1}\\ &=\answer[given]{-8} \end{align*}$ We should also examine the change in $F(x,6)$ between $x=2$ and $x=3$ : $\begin{align*} \frac{F(3,6)-F(2,6)}{3-2}&= \frac{\answer[given]{5}-\answer[given]{16}}{3-2}\\ &=\answer[given]{-11} \end{align*}$ Now if we average these values together, we see: $\eval {\pp {x} F(x,y)}_{(x,y)=(2,6)} \approx \answer [given]{-9.5}$

Let $F:\R ^2\to \R$ be a differentiable function described by the following table of values:

Estimate $F^{(0,1)}(2,6) \begin {prompt} \approx \answer {.5} \end {prompt}$

Work as we did in the example above, finding two estimates and taking their averages.

We can also estimate partial derivatives by looking at level curves.

Let $F:\R ^2\to \R$ be described by the level curves below:

The height of the level curve is marked on the curve, and we are given a point $(4,2)$ . Estimate $F^{(1,0)}(4,2)$

To estimate $F^{(1,0)}(4,2)$ we examine the change between the level curve that $\vec {p}$ is on, and the nearest level curve found by traveling on a line parallel to the $x$ -axis. Starting at $\vec {p}$ and moving to the left, we see $\begin{align*} \frac{F(4,2)-F(1,2)}{4-1}&= \frac{\answer[given]{13}-\answer[given]{7}}{4-1}\\ &=\answer[given]{2} \end{align*}$ We also should examine the change between the closest level curve on the right: $\begin{align*} \frac{F(5,2)-F(4,2)}{5-4}&= \frac{\answer[given]{17}-\answer[given]{13}}{5-4}\\ &=\answer[given]{4} \end{align*}$ Now if we average these values together, we see: $\eval {\pp {x} F(x,y)}_{(x,y)=(2,6)} \approx \answer [given]{3}$

Let $F:\R ^2\to \R$ be described by the level curves below:

The height of the level curve is marked on the curve, and we are given a point $(4,2)$ . Estimate $F^{(0,1)}(4,2) \begin {prompt} \approx \answer {-3.5} \end {prompt}$

Work as we did in the example above, finding two estimates and taking their averages.

Combining partial derivatives

While a function $f:\R \to \R$ only has one second derivative. However, functions $F:\R ^2\to R$ have $4$ second partial derivatives and functions $F:\R ^3\to \R$ have $9$ second partial derivatives! Don’t run off yet, things get better.

Let $F(x,y)$ be continuous on an open set $S$ .

The second pure partial derivative of $F$ with respect to $x$ then $x$ is $\pp {x}\left (\pp [F]{x}\right ) = \frac {\partial ^2 F}{\partial x^2} = \left (F^{(1,0)}\right )^{(1,0)}= F^{(2,0)}$
The second pure partial derivative of $F$ with respect to $x$ then $y$ is $\pp {y}\left (\pp [F]{x}\right ) = \frac {\partial ^2F}{\partial y\partial x} = \left (F^{(1,0)}\right )^{(0,1)}$

Similar definitions hold for $\frac {\partial ^2F}{\partial y^2} = F^{(0,2)}$ and $\frac {\partial ^2F}{\partial x\partial y} = \left (F^{(0,1)}\right )^{(1,0)}$ . Moreover, there is also the notion of a mixed partial derivative, $\frac {\partial ^2F}{\partial y\partial x} \quad {and}\quad \frac {\partial ^2F}{\partial x\partial y}$

The notation of second partial derivatives gives some insight into the notation of the second derivative of a function of a single variable. If $y=f(x)$ , then $f''(x) = \frac {\d ^2 y}{\d x^2}$ . The “ $\d ^2y$ ” portion means “take the derivative of $y$ twice,” while “ $\d x^2$ ” means “with respect to $x$ both times.” When we only know of functions of a single variable, this latter phrase seems silly: there is only one variable to take the derivative with respect to. Now that we understand functions of multiple variables, we see the importance of specifying which variables we are referring to.

Consider: $F(x,y) = x^3y^2 + 2xy^3+\cos (x)$ Find six first and second partial derivatives. $\begin{align*} \pp[F]{x} &= \answer{3x^2y^2+2y^3-\sin(x)}\\ \pp[F]{y} &= \answer{2x^3y+6xy^2}\\ \frac{\partial^2F}{\partial x^2} &= \answer{6xy^2-\cos(x)}\\ \frac{\partial^2F}{\partial y^2} &= \answer{2x^3+12xy}\\ \frac{\partial^2F}{\partial y\partial x} &= \answer{6x^2y+6y^2}\\ \frac{\partial^2F}{\partial x\partial y} &= \answer{6x^2y+6y^2} \end{align*}$

Notice how above $\frac {\partial ^2F}{\partial y\partial x}=\frac {\partial ^2F}{\partial x\partial y}$ . The next theorem states that it is not a coincidence.

Mixed Partial Derivatives Let $F:\R ^2\to \R$ be a function where $\frac {\partial ^2F}{\partial y\partial x}\quad \text {and}\quad \frac {\partial ^2F}{\partial x\partial y}$ are continuous on an open set $S$ . Then for each point $(x,y)$ in $S$ , $\frac {\partial ^2F}{\partial y\partial x}=\frac {\partial ^2F}{\partial x\partial y}$ . A similar result is true for functions $F:\R ^n\to \R$ .

Finding $\frac {\partial ^2F}{\partial y\partial x}$ and $\frac {\partial ^2F}{\partial x\partial y}$ independently and comparing the results provides a convenient way of checking our work.

Differentiability

In the past you may have learned

Given a function $f$ and a number $a$ in the domain of $f$ , if one can “zoom in” on the graph at $(a, f(a))$ sufficiently so that it appears to be a straight line, then the function is differentiable, and that line is the tangent line to $f(x)$ at the point $(a,f(a))$ .

We illustrate this informal definition with the following diagram:

When working with functions of several variables, intuitively, a function $F:\R ^2\to R$ is differentiable when one can “zoom-in” and the surface determined by $F$ looks like a plane.

Given a function $F:\R ^2\to \R$ and a vector $\vec {a}$ in the domain of $F$ , if one can “zoom in” on the graph at $(\vec {a}, F(\vec {a}))$ sufficiently so that it appears to be a plane, then the function is differentiable, and that plane is the tangent plane to $F$ at the point $(\vec {a},F(\vec {a}))$ .

The following theorem states that differentiable functions are continuous, followed by another theorem that provides a more tangible way of determining whether a great number of function are differentiable or not.

Differentiability Implies Continuity Let $F:\R ^n\to \R$ be defined on an open set $S$ containing $\vec {a}$ . If $F$ is differentiable at $\vec {a}$ , then $F$ is continuous at $\vec {a}$ .

A Criterion for Differentiability Let $F:\R ^n\to \R$ be defined on an open set $S$ containing $\vec {a}$ . If $\pp [F]{x_1},\pp [F]{x_2},\dots ,\pp [F]{x_n}$ are all continuous on $S$ , then $F$ is differentiable on $S$ .

The theorems assure us that essentially all functions that we see in the course of our studies here are differentiable (and hence continuous) on their natural domains. There is a small caveat to the Theorem above: it is possible for a function $F$ to be differentiable yet some partials may not be continuous. Such strange behavior of functions is a source of delight for many mathematicians.

Finding tangent planes

In your earlier calculus courses, you often found tangent lines to curves. To do this, you were given a function $f$ , a point $x=a$ , and then you produced $y = f'(a)\cdot (x-a) + f(a)$ as your tangent line. Now, with functions $F:\R ^2\to \R$ , and a point $(a,b)$ we use partial derivatives to find tangent planes. These are planes of the form: $z= F^{(1,0)}(a,b)\cdot (x-a) + F^{(0,1)}(a,b)\cdot (y-b)+F(a,b)$

Find a tangent plane to $F(x,y) = 3\cos (x)\sin (y)$ at $(x,y) = (\pi /3,\pi /6)$ . $z = \answer {(-3\sqrt {3}/4)\cdot (x-\pi /3) + (3\sqrt {3}/4)\cdot (y-\pi /6) + 3/4}$

We can also find tangent planes of surfaces that are defined parameterically.

Consider the parameterization of the unit sphere, centered at the origin where $0\le \theta < 2\pi$ and $0\le \phi \le \pi$ : $\begin{align*} x(\theta,\phi) &= \cos(\theta)\sin(\phi)\\ y(\theta,\phi) &= \sin(\theta)\sin(\phi)\\ z(\theta,\phi) &= \cos(\phi) \end{align*}$ Find a plane tangent to the sphere when $\theta = \pi /4$ and $\phi = \pi /3$ .

Here we’ll use the parametric formula for a plane: $\vec {L}(s,t) = \vec {p}+ s \vec {v} + t\cdot \vec {w}$ The the point $\vec {p}$ (denoted by a vector) is: $\vec {p} = \vector {\answer [given]{\sqrt {3/8}},\answer [given]{\sqrt {3/8}},\answer [given]{1/2}}$ The vector $\vec {v}$ is given by $\begin{align*} \vec{v} &=\eval{\pp{\theta}\vector{x(\theta,\phi),y(\theta,\phi),z(\theta,\phi)}}_{\substack{\theta = \pi/4\\ \phi = \pi/3}}\\ &= \eval{\vector{\answer[given]{-\sin(\theta)\sin(\phi)},\answer[given]{\cos(\theta)\sin(\phi)},\answer[given]{0}}}_{\substack{\theta = \pi/4\\ \phi = \pi/3}}\\ &= \vector{-\sqrt{3/8},\sqrt{3/8},0} \end{align*}$ And vector $\vec {w}$ is given by $\begin{align*} \vec{w} &=\eval{\pp{\phi}\vector{x(\theta,\phi),y(\theta,\phi),z(\theta,\phi)}}_{\substack{\theta = \pi/4\\ \phi = \pi/3}}\\ &= \eval{\vector{\answer[given]{\cos(\theta)\cos(\phi)},\answer[given]{\sin(\theta)\cos(\phi)},\answer[given]{-\sin(\phi)}}}_{\substack{\theta = \pi/4\\ \phi = \pi/3}}\\ &= \vector{1/\sqrt{8},1/\sqrt{8},-\sqrt{3}/2} \end{align*}$ Hence our desired plane is given by: $\vec {L}(s,t) = \begin {bmatrix} \answer [given]{\sqrt {3/8}-s\sqrt {3/8}+ t/\sqrt {8}}\\ \answer [given]{\sqrt {3/8} +s\sqrt {3/8} + t/\sqrt {8}}\\ \answer [given]{1/2 -t\sqrt {3}/2} \end {bmatrix}$