Total Derivative, Tangent Planes, and Differentials, multivariable and vector calculus lesson

Zoom in far enough on a smooth hill and the ground under your feet looks flat. That flat piece is the tangent plane, and the whole idea of a derivative in several variables is that the function looks like a linear map plus an error that shrinks faster than the step you take. Partial derivatives only test the function along the two axis directions, so they cannot by themselves tell you whether such a plane exists.

Key terms

Differentiability

Definition Let $U \subseteq \mathbb{R}^n$ be open, let $f \colon U \to \mathbb{R}^m$ , and let $\mathbf{a}$ be a point of $U$ . The function $f$ is differentiable at $\mathbf{a}$ if there is a linear map $L \colon \mathbb{R}^n \to \mathbb{R}^m$ such that

\begin{gathered} \lim_{\mathbf{h} \to \mathbf{0}} \frac{\lVert R(\mathbf{h}) \rVert}{\lVert \mathbf{h} \rVert} = 0, \quad \text{where} \\[0.4em] R(\mathbf{h}) = f(\mathbf{a} + \mathbf{h}) - f(\mathbf{a}) - L(\mathbf{h}). \end{gathered}

Equivalently, the remainder $R(\mathbf{h})$ is $o(\lVert \mathbf{h} \rVert)$ . Such an $L$ is unique when it exists. It is called the total derivative of $f$ at $\mathbf{a}$ and written $Df(\mathbf{a})$ . The limit is a genuine multivariable limit, so $\mathbf{h}$ may approach $\mathbf{0}$ from every direction and along every path. A function differentiable at $\mathbf{a}$ is continuous at $\mathbf{a}$ .

In plain words The change in $f$ is a linear part plus an error, and the error becomes small even compared with the size of the step. Being small is not enough, since every continuous function has an error that goes to zero. It has to vanish faster than $\lVert \mathbf{h} \rVert$ .

Example Take $f(x, y) = xy$ at $(1, 2)$ with step $(h, k)$ . Expanding gives the exact change

\begin{aligned} f(1 + h, 2 + k) &- f(1, 2) \\[0.4em] &= 2h + k + hk. \end{aligned}

The candidate linear map is $L(h, k) = 2h + k$ and the exact remainder is $R = hk$ . Since $\lvert hk \rvert \le \frac{1}{2}(h^2 + k^2)$ ,

\frac{\lvert hk \rvert}{\sqrt{h^2 + k^2}} \le \frac{1}{2}\sqrt{h^2 + k^2} \to 0,

so $f$ is differentiable at $(1, 2)$ with $Df(1, 2)(h, k) = 2h + k$ .

Jacobian matrix

Definition Write $f = (f_1, \dots, f_m)$ . If $f$ is differentiable at $\mathbf{a}$ , then every partial derivative $\frac{\partial f_i}{\partial x_j}(\mathbf{a})$ exists, and $Df(\mathbf{a})$ is multiplication by the $m \times n$ Jacobian matrix

Jf(\mathbf{a}) = \left[ \frac{\partial f_i}{\partial x_j}(\mathbf{a}) \right],

whose row $i$ holds the partial derivatives of the component $f_i$ and whose column $j$ holds the partial derivatives with respect to $x_j$ . The matrix can be written down whenever the partial derivatives exist, but it represents a derivative of $f$ only when $f$ is differentiable at $\mathbf{a}$ . For a real valued $f$ (the case $m = 1$ ) the Jacobian is the single row $\left[ f_{x_1}(\mathbf{a}) \ \cdots \ f_{x_n}(\mathbf{a}) \right]$ .

In plain words Once you know $f$ is differentiable, the linear map is found by computing partial derivatives and stacking them in a matrix. Each row describes one output, each column one input.

Example For $f(x, y) = \langle x^2 y, \ x - y^3 \rangle$ , a polynomial map and so differentiable everywhere (see the last key term), the Jacobian at $(1, 2)$ is

\begin{aligned} Jf(1, 2) &= \begin{bmatrix} 2xy & x^2 \\ 1 & -3y^2 \end{bmatrix}_{(1,2)} \\[0.4em] &= \begin{bmatrix} 4 & 1 \\ 1 & -12 \end{bmatrix}. \end{aligned}

For the step $\mathbf{h} = \langle 0.01, -0.02 \rangle$ the linear prediction is $Jf(1, 2)\mathbf{h} = \langle 0.02, 0.25 \rangle$ . The exact change is $f(1.01, 1.98) - f(1, 2) = \langle 0.019798, 0.247608 \rangle$ , off by less than $0.003$ in each component.

Linearization

Definition If $f$ is differentiable at $\mathbf{a}$ , its linearization at $\mathbf{a}$ is the affine map (a constant plus the linear map $Df(\mathbf{a})$ )

L(\mathbf{x}) = f(\mathbf{a}) + Jf(\mathbf{a})(\mathbf{x} - \mathbf{a}).

For a real valued function of two variables this reads

\begin{aligned} L(x, y) = f(a, b) &+ f_x(a, b)(x - a) \\[0.4em] &+ f_y(a, b)(y - b). \end{aligned}

The approximation $f(\mathbf{x}) \approx L(\mathbf{x})$ has error $f(\mathbf{x}) - L(\mathbf{x}) = R(\mathbf{x} - \mathbf{a})$ , which is $o(\lVert \mathbf{x} - \mathbf{a} \rVert)$ .

In plain words Replace the function near $\mathbf{a}$ by the value at $\mathbf{a}$ plus the linear change. Close to $\mathbf{a}$ the error is tiny compared with the distance from $\mathbf{a}$ .

Example The distance from the origin $f(x, y) = \sqrt{x^2 + y^2}$ has continuous partial derivatives away from the origin, so it is differentiable at $(3, 4)$ . There we get $f = 5$ , $f_x = \frac{3}{5}$ and $f_y = \frac{4}{5}$ , so

\begin{aligned} L(x, y) = 5 &+ 0.6(x - 3) \\[0.4em] &+ 0.8(y - 4). \end{aligned}

Point	$f$	$L$	Error $f - L$
$(3.1, 3.9)$	$4.9819675$	$4.98$	$0.0019675$
$(3.01, 3.99)$	$4.9980196$	$4.998$	$0.0000196$

The values are rounded to seven decimal places. Shrinking the step by a factor of $10$ shrank the error by a factor of about $100$ , so the error divided by the step length still goes to $0$ .

Tangent plane

Definition Let $f \colon U \to \mathbb{R}$ with $U \subseteq \mathbb{R}^2$ open, and let $f$ be differentiable at $(a, b)$ in $U$ . The tangent plane to the graph $z = f(x, y)$ at the point $(a, b, f(a, b))$ is the graph of the linearization,

\begin{aligned} z = f(a, b) &+ f_x(a, b)(x - a) \\[0.4em] &+ f_y(a, b)(y - b). \end{aligned}

A normal vector to it is $\langle f_x(a, b), f_y(a, b), -1 \rangle$ . If $f$ is not differentiable at $(a, b)$ , the graph has no tangent plane of this form there, even when both partial derivatives exist.

In plain words The tangent plane is the plane that hugs the surface at the point, so closely that the vertical gap between them shrinks faster than your horizontal distance from the point.

Example For $f(x, y) = x^2 + 3y^2$ at $(1, 1)$ we have $f = 4$ , $f_x = 2x = 2$ and $f_y = 6y = 6$ . The tangent plane is

\begin{aligned} z &= 4 + 2(x - 1) + 6(y - 1) \\[0.4em] &= 2x + 6y - 4, \end{aligned}

with normal vector $\langle 2, 6, -1 \rangle$ . The exact vertical gap is $f - z = (x - 1)^2 + 3(y - 1)^2$ , which is at most $3$ times the squared distance to $(1, 1)$ .

Total differential

Definition For $z = f(x, y)$ differentiable at $(x, y)$ , the total differential is

dz = f_x(x, y)\, dx + f_y(x, y)\, dy,

where $dx$ and $dy$ are independent increments. It is the linear part of the change, and the actual change $\Delta z = f(x + dx, y + dy) - f(x, y)$ satisfies $\Delta z = dz + o\big(\sqrt{dx^2 + dy^2}\big)$ . The same formula with $n$ terms holds for $n$ variables.

In plain words The differential is the quick estimate of how much the output moves when each input moves a little. It is used for error estimates and small corrections.

Example A can has radius $r = 3$ cm and height $h = 10$ cm, so $V = \pi r^2 h$ . If the radius grows by $dr = 0.05$ cm and the height changes by $dh = -0.1$ cm, then

\begin{aligned} dV &= 2\pi r h\, dr + \pi r^2\, dh \\[0.4em] &= 3\pi - 0.9\pi = 2.1\pi \\[0.4em] &\approx 6.597 \text{ cm}^3. \end{aligned}

The exact change is $\pi(3.05^2 \cdot 9.9 - 90) \approx 6.581$ cubic centimeters, so the differential is off by about $0.016$ .

Partial derivatives alone are not enough

Definition The existence of all partial derivatives at $\mathbf{a}$ does not imply that $f$ is differentiable at $\mathbf{a}$ , and it does not even imply that $f$ is continuous there. To show $f$ is not differentiable at $\mathbf{a}$ when the partial derivatives exist, use the only possible candidate, the Jacobian, and show that the remainder divided by $\lVert \mathbf{h} \rVert$ does not tend to $0$ along some path.

In plain words Partial derivatives look only along the coordinate axes. The function can behave badly along a diagonal, and differentiability notices that.

Example Let $f(x, y) = \frac{x^2 y}{x^2 + y^2}$ for $(x, y) \ne (0, 0)$ and $f(0, 0) = 0$ . On both axes $f$ is $0$ , so $f_x(0, 0) = f_y(0, 0) = 0$ , and the only candidate is $L = 0$ . Along the diagonal $h = k$ with $h > 0$ ,

\frac{f(h, h) - 0}{\sqrt{2}\,h} = \frac{h/2}{\sqrt{2}\,h} = \frac{\sqrt{2}}{4} \ne 0,

so $f$ is not differentiable at the origin and its graph has no tangent plane there. The function is continuous at the origin, since $\lvert f(x, y) \rvert \le \lvert y \rvert$ . The function $\frac{xy}{x^2 + y^2}$ , set to $0$ at the origin, is worse. It has both partial derivatives equal to $0$ there but equals $\frac{1}{2}$ along the line $y = x$ , so it is not even continuous.

Continuous partial derivatives are sufficient

Definition Let $U \subseteq \mathbb{R}^n$ be open, let $f \colon U \to \mathbb{R}^m$ , and let $\mathbf{a}$ be in $U$ . Suppose there is an open ball $B \subseteq U$ centered at $\mathbf{a}$ such that every partial derivative $\frac{\partial f_i}{\partial x_j}$ exists at every point of $B$ , and suppose each of these partial derivatives is continuous at $\mathbf{a}$ . Then $f$ is differentiable at $\mathbf{a}$ . In particular, a function of class $C^1$ on $U$ , meaning all first partial derivatives exist and are continuous on $U$ , is differentiable at every point of $U$ . The condition is sufficient but not necessary.

In plain words Checking that the partial derivatives are continuous is the everyday way to prove differentiability. Polynomials, exponentials, sines and their sums, products, quotients with nonzero denominators and compositions all pass this test on open sets where they are defined.

Example The partial derivatives of $f(x, y) = e^x \cos y$ are $e^x \cos y$ and $-e^x \sin y$ , both continuous on $\mathbb{R}^2$ . So $f$ is differentiable everywhere, and at $(0, 0)$ its linearization is $L(x, y) = 1 + x$ . Continuous partial derivatives are not necessary, though. Let $r = \sqrt{x^2 + y^2}$ and

g(x, y) = \begin{cases} r^2 \sin(1/r), & r \ne 0, \\ 0, & r = 0. \end{cases}

Since $\lvert g \rvert \le r^2$ , the remainder ratio $\lvert g(\mathbf{h}) \rvert / \lVert \mathbf{h} \rVert$ is at most $\lVert \mathbf{h} \rVert$ , so $g$ is differentiable at the origin with derivative $0$ . In particular $g_x(0, 0) = 0$ . Away from the origin the product and chain rules give

g_x = 2x \sin(1/r) - \frac{x}{r}\cos(1/r).

On the positive $x$ axis this is $2x\sin(1/x) - \cos(1/x)$ . The first term tends to $0$ as $x \to 0^+$ , while $\cos(1/x)$ takes the values $1$ and $-1$ at points arbitrarily close to $0$ , so $g_x$ has no limit at the origin and is not continuous there. By symmetry the same holds for $g_y$ .

Check yourself

Find the linearization of $f(x, y) = x e^y$ at $(2, 0)$ and use it to estimate $f(2.1, -0.1)$ .

Show the answer

The partial derivatives are $f_x = e^y = 1$ and $f_y = x e^y = 2$ at $(2, 0)$ , and they are continuous everywhere. So

L(x, y) = 2 + (x - 2) + 2y,

and $L(2.1, -0.1) = 2 + 0.1 - 0.2 = 1.9$ . The true value is $2.1 e^{-0.1} \approx 1.90016$ .

Let $g(x, y) = \frac{x^2 y^2}{x^2 + y^2}$ for $(x, y) \ne (0, 0)$ and $g(0, 0) = 0$ . Prove that $g$ is differentiable at the origin.

Show the answer

On the axes $g$ is $0$ , so both partial derivatives at the origin are $0$ and the candidate is $L = 0$ . Since $x^2 y^2 \le \frac{1}{4}(x^2 + y^2)^2$ ,

\frac{\lvert g(h, k) \rvert}{\sqrt{h^2 + k^2}} \le \frac{1}{4}\sqrt{h^2 + k^2} \to 0.

So $g$ is differentiable at the origin with total derivative $0$ .

Show that $f(x, y) = \sqrt{\lvert xy \rvert}$ has both partial derivatives at the origin but no tangent plane there.

Show the answer

The function is $0$ on both axes, so $f_x(0, 0) = f_y(0, 0) = 0$ and the only candidate plane is $z = 0$ . Along $h = k \ne 0$ the ratio is

\frac{f(h, h)}{\sqrt{2}\,\lvert h \rvert} = \frac{\lvert h \rvert}{\sqrt{2}\,\lvert h \rvert} = \frac{1}{\sqrt{2}},

which does not tend to $0$ . So $f$ is not differentiable at the origin and has no tangent plane there.

Total Derivative, Tangent Planes, and Differentials

Key terms

Differentiability

Jacobian matrix

Linearization

Tangent plane

Total differential

Partial derivatives alone are not enough

Continuous partial derivatives are sufficient

Check yourself

How this lesson connects