# Math 141, Lesson 38: Partial Derivatives II

The lesson page, with interactive tools and R in the browser: https://tangentspace.org/lessons/38/

This lesson is marked "Updates coming": parts of it are still being written.

Reading: MMAC Ch 4.6, pp 451-457

Practice: Optional HW 38 **[Q?? and Q??]**.

## Objectives

1. Using R, compute partial derivatives of a multivariable function.
2. Obtain higher-order partial derivatives by hand.
3. Use partial derivatives to obtain a linear approximation of a multivariable function at a point.

## 1. Partial Derivatives in R

*Objective.* Using R, compute partial derivatives of a multivariable function.

The R command *`D(f(x,y) ~ x)`* returns $f_x$, the partial derivative of $f(x,y)$ with respect to the variable written immediately after the tilde, as a function of every input. The command `D(f(x,y) ~ y)` returns $f_y$.

After `dfx = D(f(x,y) ~ x)`, evaluate $f_x$ at $(1,2)$ with `dfx(x = 1, y = 2)`, or with `dfx(1, 2)`, which takes the values in the order $x$, $y$. A call with an input left out stops with an error:

```r
> dfx(1)
Error in dfx(1) : argument "y" is missing, with no default
```

### Example 1

Use R to compute $f_x$ and $f_y$, then evaluate both at $(1, 2)$:

$$f(x,y) = xy^2 + x^3.$$

**Solution.**

Define $f$ with `makeFun()`, compute each first partial derivative with `D()`, and evaluate both partial derivatives at $(1,2)$. Used R.

```r
f = makeFun(x*y^2 + x^3 ~ x & y)
dfx = D(f(x,y) ~ x)
dfy = D(f(x,y) ~ y)
dfx(x = 1, y = 2)
#> [1] 7
dfy(x = 1, y = 2)
#> [1] 4
```

R returns $f_x(1,2) = 7$ and $f_y(1,2) = 4$. As a check, the command `dfx` alone prints the formula R found, `3 * x^2 + y^2`, which is $f_x(x,y) = 3x^2 + y^2$ from the derivative rules. The command `dfy` alone prints `2 * x * y`, which is $f_y(x,y) = 2xy$.

### Example 2

Use R to compute $f_x(1,2)$ and $f_y(1,2)$:

$$f(x,y) = \cos(xy) + x^2y^3.$$

**Solution.**

Define $f$, compute $f_x$ and $f_y$ with `D()`, and evaluate both at $(1,2)$. Used R.

```r
f = makeFun(cos(x*y) + x^2*y^3 ~ x & y)
dfx = D(f(x,y) ~ x)
dfy = D(f(x,y) ~ y)
dfx(x = 1, y = 2)
#> [1] 14.18141
dfy(x = 1, y = 2)
#> [1] 11.0907
```

R returns $f_x(1,2) \approx 14.181$ and $f_y(1,2) \approx 11.091$.

To check by hand, treat $y$ as a constant and apply the chain rule to $\cos(xy)$:

$$\begin{align*}
f_x &= \frac{\partial}{\partial x}\left[\cos(xy) + x^2y^3\right] = -y\sin(xy) + 2xy^3,\\
f_x(1,2) &= -2\sin(2) + 2(1)(2)^3 = 16 - 2\sin(2) \approx 14.181.
\end{align*}$$

The hand value agrees with the value from R.

### Example 3

For $f(x,y) = xy^2 + x^3$, plot $f_x$ in R for $0 \le x \le 2$ with $y$ held at 1, and determine whether $f$ increases or decreases in $x$ when $y = 1$.

**Solution.**

Define $f$ and $f_x$ in R, then plot $f_x$ with $y$ held at 1. Used R.

`f = makeFun(x*y^2 + x^3 ~ x & y)`

`dfx = D(f(x,y) ~ x)`

`plotFun(dfx(x, y = 1) ~ x, xlim = range(0, 2))`

In the formula `dfx(x, y = 1) ~ x`, the argument `y = 1` fixes $y$ at 1, and `~ x` puts $x$ on the horizontal axis.

By hand, treat $y$ as a constant, differentiate, and then set $y = 1$:

$$\begin{align*}
f_x &= \frac{\partial}{\partial x}\left[xy^2 + x^3\right] = y^2 + 3x^2,\\
f_x(x,1) &= 1^2 + 3x^2 = 1 + 3x^2.
\end{align*}$$

Because $3x^2 \ge 0$, the partial derivative satisfies $f_x(x,1) \ge 1$ for every $x$, and the plot stays above the $x$-axis. The lowest point of the plot is $(0, 1)$, because $3x^2 = 0$ only at $x = 0$, where $f_x(0,1) = 1$. The function $f$ increases in $x$ when $y = 1$, because a positive rate of change means an increasing output ([Lesson 25](https://tangentspace.org/lessons/25/) Notes, section 4). As a check, the one-input function $f(x,1) = x + x^3$ rises from $f(0,1) = 0$ to $f(2,1) = 10$.

![Figure](https://tangentspace.org/lessons/fig/0eb117db6501.svg)

## 2. Higher-Order Partial Derivatives

*Objective.* Obtain higher-order partial derivatives by hand.

**Definition 1** (Higher-order partial derivatives).

A *second partial derivative* is a partial derivative of a first partial derivative. A *higher-order partial derivative* is a partial derivative of a partial derivative, and the second partial derivatives are the higher-order partial derivatives of order 2. A two-variable function $f(x,y)$ has four second partial derivatives:

$$f_{xx} = [f_x]_x, \qquad f_{xy} = [f_x]_y, \qquad f_{yx} = [f_y]_x, \qquad
f_{yy} = [f_y]_y.$$

The notation $[f_x]_y$ means the partial derivative of $f_x$ with respect to $y$. The *order* of a partial derivative is the number of differentiations it takes: $f_x$ has order 1, $f_{xy}$ has order 2, and $f_{xxy}$ has order 3.

The first partial derivative $f_x$ is neither a second partial derivative nor a higher-order partial derivative, because it is a partial derivative of $f$ itself, and $f$ is not a partial derivative. Both definitions require a partial derivative of a partial derivative, such as $f_{xx} = [f_x]_x$ or $f_{xxy} = [f_{xx}]_y$.

As in the textbook, each function in the examples is assumed to have partial derivatives of every order at the points where they are computed.

A *mixed partial* is a higher-order partial derivative that involves differentiating with respect to different input variables, such as $f_{xy}$ and $f_{yx}$.

*Order of differentiation.* In subscript notation, find $f_{xy}$ by differentiating with respect to $x$ first and then with respect to $y$. In Leibniz notation, differentiate first with respect to the rightmost variable in the denominator and work leftward:

$$\frac{\partial^2 f}{\partial y\,\partial x} = f_{xy}.$$

The exponent on $\partial$ in the numerator is the order. For repeated differentiation with respect to one variable, write an exponent on that variable in the denominator:

$$\frac{\partial^2 f}{\partial x^2} = f_{xx}, \qquad \frac{\partial^2 f}{\partial y^2} = f_{yy}.$$

A multivariable function $f(x,y)$ is *smooth* on an open region if and only if $f(x,y)$ has continuous partials of all orders at every point of that region.

The common modeling functions are smooth on every open region inside their domains.

**Theorem 2** (Clairaut’s theorem).

If $f(x,y)$ is smooth on an open region, then $f_{xy}(x,y) = f_{yx}(x,y)$ at every point of that region.

### Example 4

For a function $f(x,y,z)$, list the order of differentiation for each partial derivative:

$$f_{yzx}, \qquad\qquad \frac{\partial^3 f}{\partial z\,\partial y\,\partial x}.$$

**Solution.**

In $f_{yzx}$, the order is $y$, then $z$, then $x$, because subscripts are read left to right. In the partial derivative written in Leibniz notation, the order is $x$, then $y$, then $z$, because the Leibniz denominator is read from the rightmost variable leftward. In subscript notation, that partial derivative is $f_{xyz}$.

### Example 5

Compute all second partial derivatives of

$$f(x,y) = xy^2 + x^3.$$

**Solution.**

By the definition of higher-order partial derivatives, each second partial derivative is a partial derivative of a first one. Compute $f_x$ and differentiate it with respect to $x$ and to $y$, then do the same with $f_y$.

Differentiate $f$ with respect to $x$, treating $y$ as a constant:

$$f_x = \frac{\partial}{\partial x}\left[xy^2 + x^3\right] = y^2 + 3x^2.$$

Differentiate $f_x$ with respect to $x$ and then with respect to $y$:

$$f_{xx} = \frac{\partial}{\partial x}\left[y^2 + 3x^2\right] = 6x, \qquad
f_{xy} = \frac{\partial}{\partial y}\left[y^2 + 3x^2\right] = 2y.$$

Differentiate $f$ with respect to $y$, treating $x$ as a constant:

$$f_y = \frac{\partial}{\partial y}\left[xy^2 + x^3\right] = x(2y) + 0 = 2xy.$$

Differentiate $f_y$ with respect to $x$ and then with respect to $y$:

$$f_{yx} = \frac{\partial}{\partial x}\left[2xy\right] = 2y, \qquad
f_{yy} = \frac{\partial}{\partial y}\left[2xy\right] = 2x.$$

The mixed partials $f_{xy}$ and $f_{yx}$ both equal $2y$. Clairaut’s theorem predicts this equality, because a polynomial is smooth everywhere.

### Example 6

Compute all second partial derivatives of

$$g(x,y) = x^4\sin(y).$$

**Solution.**

As in Example 5, start with $g_x$, where the factor $\sin(y)$ is a constant multiple, and differentiate it with respect to $x$ and then $y$:

$$\begin{align*}
g_x &= \frac{\partial}{\partial x}\bigl[x^4\sin(y)\bigr] = 4x^3\sin(y),\\
g_{xx} &= \frac{\partial}{\partial x}\bigl[4x^3\sin(y)\bigr] = 12x^2\sin(y), \qquad
g_{xy} = \frac{\partial}{\partial y}\bigl[4x^3\sin(y)\bigr] = 4x^3\cos(y).
\end{align*}$$

Do the same with $g_y$, where the factor $x^4$ is a constant multiple:

$$\begin{align*}
g_y &= \frac{\partial}{\partial y}\bigl[x^4\sin(y)\bigr] = x^4\cos(y),\\
g_{yx} &= \frac{\partial}{\partial x}\bigl[x^4\cos(y)\bigr] = 4x^3\cos(y), \qquad
g_{yy} = \frac{\partial}{\partial y}\bigl[x^4\cos(y)\bigr] = -x^4\sin(y).
\end{align*}$$

The mixed partials $g_{xy}$ and $g_{yx}$ are equal, as Clairaut’s theorem states for the smooth function $x^4\sin(y)$.

### Example 7

Compute all second partial derivatives of the function below, then compute the third partial derivative $\dfrac{\partial^3 q}{\partial y\,\partial x^2}$:

$$q(x,y) = x^3e^{4y}.$$

**Solution.**

As in Example 5, start with $q_x$, where the factor $e^{4y}$ is a constant multiple, and differentiate it with respect to $x$ and then $y$. The computation of $q_{xy}$ uses $\frac{\partial}{\partial y}\bigl[e^{4y}\bigr] = 4e^{4y}$ from the chain rule:

$$\begin{align*}
q_x &= \frac{\partial}{\partial x}\bigl[x^3e^{4y}\bigr] = 3x^2e^{4y},\\
q_{xx} &= \frac{\partial}{\partial x}\bigl[3x^2e^{4y}\bigr] = 6xe^{4y}, \qquad
q_{xy} = \frac{\partial}{\partial y}\bigl[3x^2e^{4y}\bigr] = 3x^2 \cdot 4e^{4y} = 12x^2e^{4y}.
\end{align*}$$

Do the same with $q_y$, where the factor $x^3$ is a constant multiple:

$$\begin{align*}
q_y &= \frac{\partial}{\partial y}\bigl[x^3e^{4y}\bigr] = x^3 \cdot 4e^{4y} = 4x^3e^{4y},\\
q_{yx} &= \frac{\partial}{\partial x}\bigl[4x^3e^{4y}\bigr] = 12x^2e^{4y}, \qquad
q_{yy} = \frac{\partial}{\partial y}\bigl[4x^3e^{4y}\bigr] = 4x^3 \cdot 4e^{4y} = 16x^3e^{4y}.
\end{align*}$$

The coefficient 16 of $q_{yy}$ is $4 \cdot 4$, one factor of 4 from each differentiation with respect to $y$.

The Leibniz denominator $\partial y\,\partial x^2$ read from the right gives the order $x$, $x$, then $y$. The third partial derivative is $q_{xxy}$. Differentiate $q_{xx}$ with respect to $y$:

$$q_{xxy} = \frac{\partial}{\partial y}\bigl[6xe^{4y}\bigr] = 6x \cdot 4e^{4y} = 24xe^{4y}.$$

*Higher-order partial derivatives in R.* List the variables of differentiation in order after the tilde, separated by `&`. The call `D(f(x,y) ~  
x & y)` computes $f_{xy}$. The call `D(f(x,y) ~ x & x)` computes $f_{xx}$.

### Example 8

The second partial derivatives of $g(x,y) = x^4\sin(y)$ are

$$g_{xx} = 12x^2\sin(y), \qquad g_{xy} = g_{yx} = 4x^3\cos(y), \qquad g_{yy} = -x^4\sin(y).$$

Use R to check them at the point $(2, 1)$.

**Solution.**

Define $g$, compute the four second partial derivatives with `D()` by listing the variables of differentiation in order, and evaluate each at $(2,1)$. Used R.

```r
g = makeFun(x^4*sin(y) ~ x & y)
dgxx = D(g(x,y) ~ x & x)
dgxy = D(g(x,y) ~ x & y)
dgyx = D(g(x,y) ~ y & x)
dgyy = D(g(x,y) ~ y & y)
dgxx(x = 2, y = 1)
#> [1] 40.39061
dgxy(x = 2, y = 1)
#> [1] 17.28967
dgyx(x = 2, y = 1)
#> [1] 17.28967
dgyy(x = 2, y = 1)
#> [1] -13.46354
```

The R values are $g_{xx}(2,1) \approx 40.391$, $g_{xy}(2,1) = g_{yx}(2,1) \approx 17.290$, and $g_{yy}(2,1) \approx -13.464$.

Substitute $(2,1)$ into the hand formulas. Used calculator.

$$\begin{align*}
g_{xx}(2,1) &= 12(2)^2\sin(1) = 48\sin(1) \approx 40.391,\\
g_{xy}(2,1) = g_{yx}(2,1) &= 4(2)^3\cos(1) = 32\cos(1) \approx 17.290,\\
g_{yy}(2,1) &= -(2)^4\sin(1) = -16\sin(1) \approx -13.464.
\end{align*}$$

The hand values match R at $(2,1)$, a check on the formulas.

## 3. Linear Approximation

*Objective.* Use partial derivatives to obtain a linear approximation of a multivariable function at a point.

![Figure](https://tangentspace.org/lessons/fig/83e2570e9be6.svg)

The figure shows the surface $z = f(x,y)$ for $f(x,y) = x^2 + y^2$ and, in orange, the plane tangent to the surface at the point $(1, 1, 2)$. The point lies on the surface because $f(1,1) = 1^2 + 1^2 = 2$. From $f_x = 2x$ and $f_y = 2y$, the plane’s slope is $f_x(1,1) = 2(1) = 2$ in the $x$-direction and $f_y(1,1) = 2(1) = 2$ in the $y$-direction. The plane has the equation

$$z = 2 + 2(x - 1) + 2(y - 1).$$

Around $(1,1)$, the plane and the surface nearly coincide. The plane is the graph of the linear approximation, defined next.

**Definition 3** (Linear approximation in two inputs).

Let $f(x,y)$ be a smooth function around the point $(a,b)$. The *linear approximation* of $f(x,y)$ when $(x,y) = (a,b)$ is given by

$$L(x,y) = f(a,b) + f_x(a,b)\cdot(x - a) + f_y(a,b)\cdot(y - b).$$

The function $k(x,y) = |x| + y$ has no linear approximation when $(x,y) = (0,0)$, because it is not smooth around $(0,0)$: section 2 of the [Lesson 37](https://tangentspace.org/lessons/37/) Notes shows that $k_x(0,0)$ does not exist. The definition requires a smooth function, and the formula for $L$ contains $k_x(0,0)$.

The graph of $L(x,y)$ is a plane, the two-input counterpart of the tangent line of the single-variable linear approximation in section 1 of the [Lesson 30](https://tangentspace.org/lessons/30/) Notes. For inputs near $(a,b)$, the linear approximation $L(x,y)$ estimates $f(x,y)$.

*Finding a linear approximation.*

1. Compute the first partial derivatives $f_x$ and $f_y$.
2. Evaluate $f$, $f_x$, and $f_y$ at $(a,b)$.
3. Substitute $f(a,b)$, $f_x(a,b)$, and $f_y(a,b)$ into $L(x,y)$ and leave $x$ and $y$ as variables.

### Example 9

Find the linear approximation of $f(x,y)$ when $(x,y) = (1,2)$, and use it to estimate $f(1.1, 1.9)$:

$$f(x,y) = xy + e^{xy}.$$

**Solution.**

By the definition of the linear approximation, the linear approximation at $(a,b) = (1,2)$ is

$$L(x,y) = f(1,2) + f_x(1,2)\cdot(x - 1) + f_y(1,2)\cdot(y - 2).$$

To write the approximation, find the output $f(1,2)$ and the two slopes $f_x(1,2)$ and $f_y(1,2)$.

Compute the first partial derivatives. The chain rule gives $\frac{\partial}{\partial x}\bigl[e^{xy}\bigr] = ye^{xy}$ and $\frac{\partial}{\partial y}\bigl[e^{xy}\bigr] = xe^{xy}$:

$$f_x = \frac{\partial}{\partial x}\bigl[xy + e^{xy}\bigr] = y + ye^{xy}, \qquad
f_y = \frac{\partial}{\partial y}\bigl[xy + e^{xy}\bigr] = x + xe^{xy}.$$

Evaluate $f$, $f_x$, and $f_y$ at $(1,2)$:

$$\begin{align*}
f(1,2) &= (1)(2) + e^{(1)(2)} = 2 + e^2,\\
f_x(1,2) &= 2 + 2e^{(1)(2)} = 2 + 2e^2,\\
f_y(1,2) &= 1 + 1 \cdot e^{(1)(2)} = 1 + e^2.
\end{align*}$$

Substitute the three values into the linear approximation:

$$L(x,y) = (2 + e^2) + (2 + 2e^2)(x - 1) + (1 + e^2)(y - 2).$$

Evaluate $L$ at $(1.1, 1.9)$, keeping $e^2$ exact until the last step. Used calculator.

$$\begin{align*}
L(1.1, 1.9) &= (2 + e^2) + (2 + 2e^2)(1.1 - 1) + (1 + e^2)(1.9 - 2)\\
 &= (2 + e^2) + (0.2 + 0.2e^2) + (-0.1 - 0.1e^2)\\
 &= 2.1 + 1.1e^2 \approx 10.228.
\end{align*}$$

As a check, the actual value is

$$f(1.1, 1.9) = (1.1)(1.9) + e^{(1.1)(1.9)} = 2.09 + e^{2.09} \approx 10.175,$$

below the estimate 10.228 by about 0.053.

### Example 10

Use R to build the linear approximation of $f(x,y) = xy + e^{xy}$ when $(x,y) = (1,2)$, and compute the estimate of $f(1.1, 1.9)$.

**Solution.**

Define $f$, compute $f_x$ and $f_y$ with `D()`, and evaluate all three at $(1,2)$. Used R.

```r
f = makeFun(x*y + exp(x*y) ~ x & y)
dfx = D(f(x,y) ~ x)
dfy = D(f(x,y) ~ y)
f(x = 1, y = 2)
#> [1] 9.389056
dfx(x = 1, y = 2)
#> [1] 16.77811
dfy(x = 1, y = 2)
#> [1] 8.389056
```

As a check by hand, $f_x = y + ye^{xy}$ and $f_y = x + xe^{xy}$ give $f(1,2) = 2 + e^2$, $f_x(1,2) = 2 + 2e^2$, and $f_y(1,2) = 1 + e^2$, which match R with $e^2 \approx 7.389056$. The linear approximation is

$$L(x,y) \approx 9.389 + 16.778(x - 1) + 8.389(y - 2).$$

Build $L$ from the unrounded R values and evaluate it at $(1.1, 1.9)$:

```r
L = makeFun(f(1, 2) + dfx(1, 2)*(x - 1) +
dfy(1, 2)*(y - 2) ~ x & y)
L(x = 1.1, y = 1.9)
#> [1] 10.22796
```

The estimate is $f(1.1, 1.9) \approx L(1.1, 1.9) \approx 10.228$.

### Example 11

A linear approximation of a function $f(x,y)$ is

$$L(x,y) = 5 - 2(x - 3) + 4(y + 1).$$

Determine the point $(a,b)$ at which $L$ approximates $f$, the value of $f$ at that point, and whether $f$ increases or decreases in each input near that point.

**Solution.**

Rewrite $y + 1$ as $y - (-1)$ and match the terms to the linear approximation formula:

$$L(x,y) = 5 + (-2)\cdot(x - 3) + 4\cdot\bigl(y - (-1)\bigr)
       = f(a,b) + f_x(a,b)\cdot(x - a) + f_y(a,b)\cdot(y - b).$$

The match gives $a = 3$ and $b = -1$. The linear approximation is taken at $(3, -1)$, where $f(3,-1) = 5$. Near that point, $f$ decreases in $x$ because $f_x(3,-1) = -2$ is negative. It increases in $y$ because $f_y(3,-1) = 4$ is positive.

### Example 12

A cylinder with radius $r$ centimeters and height $h$ centimeters has volume

$$\text{Vol}(r,h) = \pi r^2 h$$

cubic centimeters. Find the linear approximation of Vol when $(r,h) = (2,5)$, and use it to estimate the volume when the radius is 2.1 centimeters and the height is 4.9 centimeters.

**Solution.**

To write the linear approximation, find $\text{Vol}(2,5)$, $\text{Vol}_r(2,5)$, and $\text{Vol}_h(2,5)$. Compute the first partial derivatives. With respect to $r$, the factor $\pi h$ is a constant multiple. With respect to $h$, the factor $\pi r^2$ is a constant multiple:

$$\text{Vol}_r = \frac{\partial}{\partial r}\bigl[\pi r^2 h\bigr] = 2\pi rh, \qquad
\text{Vol}_h = \frac{\partial}{\partial h}\bigl[\pi r^2 h\bigr] = \pi r^2.$$

Evaluate all three at $(2,5)$:

$$\begin{align*}
\text{Vol}(2,5) &= \pi(2)^2(5) = 20\pi,\\
\text{Vol}_r(2,5) &= 2\pi(2)(5) = 20\pi,\\
\text{Vol}_h(2,5) &= \pi(2)^2 = 4\pi.
\end{align*}$$

Substitute the three values into the linear approximation:

$$L(r,h) = 20\pi + 20\pi(r - 2) + 4\pi(h - 5).$$

Evaluate $L$ at $(2.1, 4.9)$. Used calculator.

$$\begin{align*}
L(2.1, 4.9) &= 20\pi + 20\pi(2.1 - 2) + 4\pi(4.9 - 5)\\
 &= 20\pi + 2\pi - 0.4\pi = 21.6\pi \approx 67.858.
\end{align*}$$

The estimated volume is about 67.858 cubic centimeters. As a check, the actual volume is

$$\text{Vol}(2.1, 4.9) = \pi(2.1)^2(4.9) = 21.609\pi \approx 67.887$$

cubic centimeters. The actual volume is above the estimate $21.6\pi$ by $0.009\pi \approx 0.028$ cubic centimeters.