The Remainder Definition of Differentiability

Why the Error Must Be Measured Relative to Distance

A function can be approximated near a point by a constant plus a linear expression in the displacement. The constant is the function's value at the point; the linear expression describes the first-order change. The definition of differentiability makes this approximation precise by requiring that its error become negligible compared with the distance moved.

This comparison matters because the error need not be exactly zero. For instance, a quadratic error can remain present while still becoming small faster than the displacement itself. Differentiability captures this kind of first-order accuracy without requiring the function to be exactly linear near the point.

The Remainder Definition

Let \(U\) be an open subset of \(\mathbb{R}^n\), let \(f:U\to\mathbb{R}\), and let \(\mathbf{a}\in U\). The displacement from \(\mathbf{a}\) is a vector \(\mathbf{h}\) for which \(\mathbf{a}+\mathbf{h}\in U\). Use the Euclidean norm \(\|\mathbf{h}\|=\sqrt{h_1^2+\cdots+h_n^2}\) to measure the length of that displacement.

Definition: The function \(f\) is differentiable at \(\mathbf{a}\) if there is a linear map \(L:\mathbb{R}^n\to\mathbb{R}\) and a remainder function \(r\) such that \(f(\mathbf{a}+\mathbf{h})=f(\mathbf{a})+L(\mathbf{h})+r(\mathbf{h})\) for all sufficiently small \(\mathbf{h}\), and \(\lim_{\mathbf{h}\to\mathbf{0},\,\mathbf{h}\ne\mathbf{0}} r(\mathbf{h})/\|\mathbf{h}\|=0\).

The map \(L\) is the first-order, or linear, part of the change in \(f\). The remainder \(r(\mathbf{h})\) is what is left after that linear part is subtracted. The limit says that for every positive tolerance, the ratio \(|r(\mathbf{h})|/\|\mathbf{h}\|\) is below that tolerance whenever \(\mathbf{h}\) is sufficiently small and nonzero. The remainder may be nonzero; its size relative to the displacement must tend to zero.

A linear map \(L:\mathbb{R}^n\to\mathbb{R}\) is a map satisfying \(L(\mathbf{u}+\mathbf{v})=L(\mathbf{u})+L(\mathbf{v})\) and \(L(c\mathbf{u})=cL(\mathbf{u})\) for vectors \(\mathbf{u},\mathbf{v}\) and scalar \(c\). In coordinates, every such map has the form \(L(\mathbf{h})=c_1h_1+\cdots+c_nh_n\) for fixed coefficients \(c_1,\ldots,c_n\). Thus, for a scalar function, the possible first-order approximation is a linear functional: it takes a displacement vector as input and returns a scalar.

How the Coordinate Partial Derivatives Determine the Map

When \(f\) is differentiable at \(\mathbf{a}\), the linear map in the definition is not an arbitrary choice. Its coefficient in each coordinate direction must equal the corresponding partial derivative. This follows directly from the remainder condition.

Theorem: If \(f\) is differentiable at \(\mathbf{a}=(a_1,\ldots,a_n)\), then every coordinate partial derivative \(\frac{\partial f}{\partial x_i}(\mathbf{a})\) exists, and the linear map in the definition is \(L(\mathbf{h})=\sum_{i=1}^{n}\frac{\partial f}{\partial x_i}(\mathbf{a})h_i\).

To justify the result, let \(\mathbf{e}_i\) be the vector with \(1\) in coordinate \(i\) and \(0\) in all other coordinates. Put \(\mathbf{h}=t\mathbf{e}_i\) in the defining expansion. Since \(\|t\mathbf{e}_i\|=|t|\) and \(L(t\mathbf{e}_i)=tL(\mathbf{e}_i)\), we get

\frac{f(\mathbf{a}+t\mathbf{e}_i)-f(\mathbf{a})}{t} = L(\mathbf{e}_i)+\frac{r(t\mathbf{e}_i)}{t}.

For nonzero \(t\), the absolute value of the last term is

\left|\frac{r(t\mathbf{e}_i)}{t}\right| = \frac{|r(t\mathbf{e}_i)|}{\|t\mathbf{e}_i\|} \longrightarrow 0 \qquad\text{as }t\to 0.

Therefore the coordinate difference quotient has a limit, so the partial derivative exists and equals \(L(\mathbf{e}_i)\). A linear map is determined by its values on the coordinate vectors: if \(\mathbf{h}=\sum_{i=1}^n h_i\mathbf{e}_i\), linearity gives \(L(\mathbf{h})=\sum_{i=1}^n h_iL(\mathbf{e}_i)\). Substituting the partial derivatives for those values gives the stated formula.

This also explains how to identify the only possible map: find the coordinate partial derivatives at the point and use them as the coefficients. The definition still requires the remainder condition. Knowing the coefficients identifies the candidate linear map; the ratio limit is what establishes differentiability.

Worked Examples

Worked Example: A One-Variable Remainder

Let \(f(t)=t^2\), and examine differentiability at \(a=3\). Write a nearby input as \(3+h\). Direct expansion gives

f(3+h)=(3+h)^2=9+6h+h^2, \qquad f(3)=9.

The linear term in \(h\) suggests \(L(h)=6h\), and the remainder is \(r(h)=h^2\). For \(h\ne 0\),

\frac{|r(h)|}{|h|} = \frac{|h^2|}{|h|} = |h| \longrightarrow 0 \qquad\text{as }h\to 0.

Thus \(f(3+h)=f(3)+L(h)+r(h)\) satisfies the definition. As a check, the one-variable derivative is \(f'(t)=2t\), so \(f'(3)=6\), matching the coefficient in \(L\).

Worked Example: A Two-Variable Quadratic

Let \(f(x,y)=x^2+3xy-y^2\) at \(\mathbf{a}=(1,2)\). First, evaluate the function at the point:

$$ f(1,2)=1^2+3(1)(2)-2^2=1+6-4=3. $$

Write the displacement as \(\mathbf{h}=(h,k)\). The coordinate partial derivatives are \(\frac{\partial f}{\partial x}=2x+3y\) and \(\frac{\partial f}{\partial y}=3x-2y\). At \((1,2)\), these are \(2+6=8\) and \(3-4=-1\), respectively. The candidate linear map is therefore \(L(h,k)=8h-k\).

Now expand the function at the displaced point:

\begin{aligned} f(1+h,2+k) &=(1+h)^2+3(1+h)(2+k)-(2+k)^2\\ &=(1+2h+h^2)+(6+6h+3k+3hk)-(4+4k+k^2)\\ &=3+8h-k+h^2+3hk-k^2. \end{aligned}

Thus \(r(h,k)=h^2+3hk-k^2\). Let \(\rho=\sqrt{h^2+k^2}=\|\mathbf{h}\|\). Since \(2|hk|\le h^2+k^2=\rho^2\), we have

|r(h,k)| \le h^2+3|hk|+k^2 \le \rho^2+\frac{3}{2}\rho^2 =\frac{5}{2}\rho^2.

For \(\rho>0\), this gives \(|r(h,k)|/\rho\le (5/2)\rho\), which tends to zero as \(\mathbf{h}\to\mathbf{0}\). The proposed map satisfies the remainder definition. Substituting \(h=k=0\) into the expansion also recovers \(f(1,2)=3\), as required.

Worked Example: A Linear Map in Three Variables

Consider \(f(x,y,z)=3x-2y+z+xy\) at \(\mathbf{a}=(1,2,-1)\). The function value is

$$ f(1,2,-1)=3(1)-2(2)+(-1)+(1)(2)=3-4-1+2=0. $$

The partial derivatives are \(\frac{\partial f}{\partial x}=3+y\), \(\frac{\partial f}{\partial y}=-2+x\), and \(\frac{\partial f}{\partial z}=1\). Their values at \(\mathbf{a}\) are \(5\), \(-1\), and \(1\), so the candidate map is \(L(h,k,\ell)=5h-k+\ell\).

Substitute \((1+h,2+k,-1+\ell)\) into \(f\) and expand each term:

\begin{aligned} f(1+h,2+k,-1+\ell) &=3(1+h)-2(2+k)+(-1+\ell)+(1+h)(2+k)\\ &=(3+3h)+(-4-2k)+(-1+\ell)+(2+k+2h+hk)\\ &=5h-k+\ell+hk. \end{aligned}

Because \(f(\mathbf{a})=0\), the remainder is \(r(h,k,\ell)=hk\). With \(\rho=\sqrt{h^2+k^2+\ell^2}\), the inequality \(2|hk|\le h^2+k^2\le\rho^2\) yields

\frac{|r(h,k,\ell)|}{\rho} = \frac{|hk|}{\rho} \le \frac{\rho}{2} \longrightarrow 0 \qquad\text{as }\rho\to 0,\quad \rho>0.

Hence this linear map gives the required first-order approximation. Setting the displacement to zero in the expansion gives \(f(\mathbf{a})=0\), and the coefficients of \(h,k,\ell\) agree with the three partial derivatives already calculated.

A Common Pitfall: Using the Wrong Distance

The denominator in the remainder ratio must measure the length of the whole displacement vector. Replacing it with just one coordinate can produce a quotient that is undefined or misleading. For example, let \(f(x,y)=x^2+y^2\) at the origin. The zero linear map gives the exact remainder \(r(h,k)=h^2+k^2\).

A mistaken calculation divides by \(|h|\) and writes \((h^2+k^2)/|h|\). This expression is undefined whenever \(h=0\), including for nonzero displacements such as \((0,k)\). It is not the ratio required by the definition. The correct denominator is the Euclidean distance \(\rho=\sqrt{h^2+k^2}\), so

\frac{|r(h,k)|}{\|(h,k)\|} = \frac{h^2+k^2}{\sqrt{h^2+k^2}} = \sqrt{h^2+k^2} \longrightarrow 0 \qquad\text{as }(h,k)\to(0,0),\quad (h,k)\ne(0,0).

This calculation uses the full displacement and verifies the required limit. Checking only selected directions, or using a denominator that ignores a coordinate, does not establish the remainder condition.

What the Definition Emphasizes

The definition separates differentiability into two ingredients. First, there must be a linear map that captures the first-order change. Second, after subtracting that map, the remaining error must be negligible compared with the distance to the point. Both ingredients matter: a linear expression alone is only a proposed approximation, while the limit condition measures how accurate it is nearby.

For a scalar function, the coordinate partial derivatives specify the coefficients of the only possible linear map. In two variables, for example, it has the form \(L(h,k)=f_x(a,b)h+f_y(a,b)k\). In three variables, it has the form \(L(h,k,\ell)=f_x(a,b,c)h+f_y(a,b,c)k+f_z(a,b,c)\ell\). In either case, one must still check that the remainder divided by the displacement length tends to zero. This is the defining test of first-order approximation, not merely a formula for guessing its linear part.

Check Your Understanding

Use the remainder definition and the role of the linear map to answer these questions.

In the expansion \(f(\mathbf{a}+\mathbf{h})=f(\mathbf{a})+L(\mathbf{h})+r(\mathbf{h})\), what limit must the remainder satisfy for differentiability?
Why does the value \(L(\mathbf{e}_i)\) equal the \(i\)-th coordinate partial derivative when \(f\) is differentiable?
For \(f(x,y)=x^2+y^2\) at \((0,0)\), what linear map and remainder give an expansion of the required form?
Why is dividing a multivariable remainder by just one coordinate not the definition's distance comparison?
What information do the coordinate partial derivatives provide, and what additional condition must still be checked?

The Remainder Definition of Differentiability

What You'll Learn

Why the Error Must Be Measured Relative to Distance