Derivative as Local Linear Approximation

The Derivative as a Local Model

In “Geometric Meaning of the Derivative,” the derivative supplied the slope of the tangent line at a point. That line does more than give a geometric direction: near the point of tangency, it approximates the graph. The derivative measures the coefficient of the leading linear change in the function, while the difference between the function and its tangent line becomes small compared with the distance from the point.

Let $I$ be an interval, let $a$ be an interior point of $I$, and suppose $f:I\to\mathbb{R}$ is differentiable at $a$. Write $h=x-a$, so that $x=a+h$. The tangent line at $(a,f(a))$, evaluated at $a+h$, has height $f(a)+f'(a)h$. Thus the natural comparison is between $f(a+h)$ and this linear expression.

Definition: The local linear approximation to $f$ at $a$ is the function of $h$ given by $$ f(a)+f'(a)h. $$ The remainder after this approximation is $$ R(h)=f(a+h)-f(a)-f'(a)h. $$ Differentiability at $a$ means that the remainder is small relative to $|h|$: as $h\to0$, $h\ne0$, the ratio $|R(h)|/|h|$ tends to zero.

This statement follows from the First-Order Approximation Characterization established in “Definition of the Derivative.” In the language of the difference quotient, subtracting $f'(a)$ from that quotient gives $R(h)/h$. The characterization says this difference tends to zero. Taking absolute values gives the equivalent error description above.

f(a+h)=f(a)+f'(a)h+R(h), \qquad \lim_{h\to0,\ h\ne0}\frac{|R(h)|}{|h|}=0.

The limit describes a relative scale, not necessarily a small absolute error for every chosen $h$. More precisely, for every $\varepsilon>0$, there is a $\delta>0$ such that whenever $0<|h|<\delta$ and $a+h\in I$,

|f(a+h)-f(a)-f'(a)h|<\varepsilon |h|.

The tangent-line estimate is therefore controlled by the size of the input change. For a sufficiently small input change, the error is less than any prescribed positive multiple of that change. The approximation need not be exact, and the error need not be zero; the defining feature is that its size, divided by $|h|$, tends to zero.

Computing the Approximation and Its Error

Worked Example: A Quartic Near One

Let $f(x)=x^4$ and $a=1$. For $h\ne0$, the difference quotient is

\frac{f(1+h)-f(1)}{h} =\frac{(1+h)^4-1}{h} =\frac{4h+6h^2+4h^3+h^4}{h} =4+6h+4h^2+h^3.

Its limit is $4$, so $f'(1)=4$, and the local linear approximation is $1+4h$. The exact value of the function is

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

Consequently, the remainder is $R(h)=6h^2+4h^3+h^4$. Dividing its absolute value by $|h|$ gives a quantity bounded by $6|h|+4|h|^2+|h|^3$, which tends to zero as $h\to0$. The line $1+4h$ captures the first-order change; the remaining terms are smaller relative to $|h|$.

Worked Example: A Rational Function Near Zero

Let $f(x)=1/(2+x)$ on an interval around $0$ that does not contain $-2$. Here $f(0)=1/2$. For $h\ne0$,

\frac{f(h)-f(0)}{h} =\frac{\frac{1}{2+h}-\frac12}{h} =\frac{-h}{2(2+h)h} =-\frac{1}{2(2+h)}.

As $h\to0$, this quotient tends to $-1/4$. Thus the local linear approximation is $1/2-h/4$. To find the exact remainder, calculate

R(h)=\frac{1}{2+h}-\frac12+\frac{h}{4} =\frac{4-2(2+h)+h(2+h)}{4(2+h)} =\frac{h^2}{4(2+h)}.

The numerator in the common-denominator expression simplifies as $4-4-2h+2h+h^2=h^2$. For $h$ sufficiently close to zero, $2+h$ is bounded away from zero, and therefore

\frac{|R(h)|}{|h|} =\frac{|h|}{4|2+h|} \longrightarrow 0.

This verifies directly that the linear approximation has an error small relative to the input change.

Worked Example: The Square Root Near Four

Let $f(x)=\sqrt{x}$ for $x\ge0$, and take $a=4$. For small nonzero $h$ with $4+h\ge0$, rationalizing gives

\frac{\sqrt{4+h}-2}{h} =\frac{1}{\sqrt{4+h}+2}.

By continuity of the square-root function, the quotient tends to $1/4$, so $f'(4)=1/4$. The approximation is $2+h/4$. Its remainder can also be computed exactly:

\begin{aligned} R(h) &=\sqrt{4+h}-2-\frac{h}{4}\\ &=\frac{(4+h)-(2+h/4)^2}{\sqrt{4+h}+2+h/4}\\ &=-\frac{h^2}{16\bigl(\sqrt{4+h}+2+h/4\bigr)}. \end{aligned}

The denominator tends to $64$, so it stays positive and bounded away from zero for all sufficiently small $h$. It follows that $|R(h)|/|h|$ tends to zero. In this case the error is of size proportional to $h^2$ near zero, which is smaller than a quantity proportional to $|h|$.

What a Nonzero Derivative Predicts

The local linear approximation gives a direct conclusion about whether nearby function values lie above or below $f(a)$. The conclusion uses the sign of $f'(a)$, but applies only to values compared with $f(a)$; it does not by itself show that the function is monotone throughout a neighborhood.

Theorem (Local Sign from a Nonzero Derivative): Suppose $f$ is differentiable at an interior point $a$, and $f'(a)\ne0$. If $f'(a)>0$, then for all sufficiently small $h>0$, $f(a+h)>f(a)$, and for all sufficiently small $h<0$, $f(a+h)<f(a)$. If $f'(a)<0$, both inequalities are reversed.

Proof. Put $m=f'(a)$, and use the remainder $R(h)$ from the local linear approximation. Since $|R(h)|/|h|\to0$, there is a $\delta>0$ such that $0<|h|<\delta$ implies $|R(h)|<(|m|/2)|h|$. We also restrict $h$ so that $a+h\in I$. Then

$$ f(a+h)-f(a)=mh+R(h). $$

First suppose $m>0$. If $0<h<\delta$, then $R(h)>-(m/2)h$, and hence

f(a+h)-f(a)>mh-\frac{m}{2}h=\frac{m}{2}h>0.

If $-\delta<h<0$, then $R(h)<(m/2)|h|=-(m/2)h$, so

f(a+h)-f(a)<mh-\frac{m}{2}h=\frac{m}{2}h<0.

Now suppose $m<0$. Write $c=|m|=-m>0$. For $0<h<\delta$,

f(a+h)-f(a)= -ch+R(h) < -ch+\frac{c}{2}h = -\frac{c}{2}h<0.

For $-\delta<h<0$, the remainder bound gives $R(h)>-(c/2)|h|=(c/2)h$. Therefore

f(a+h)-f(a)>-ch+\frac{c}{2}h =-\frac{c}{2}h>0.

These are exactly the claimed inequalities in both sign cases. $\square$

Worked Example: The Approximation Predicts Which Side Is Higher

For $f(x)=x^4$ at $a=1$, the derivative is $4>0$. The theorem says that sufficiently close points to the right have values greater than $f(1)$, while sufficiently close points to the left have values less than $f(1)$. Here this can be checked exactly: for $h$ close to zero,

f(1+h)-f(1)=4h+6h^2+4h^3+h^4 =h\bigl(4+6h+4h^2+h^3\bigr).

The factor in parentheses tends to $4$, so it is positive for all sufficiently small $h$. The sign of the entire expression is then the sign of $h$, as the theorem predicts. This comparison concerns the value at $1$; the theorem has not asserted that $x^4$ is monotone on a whole interval.

First-Order Contact Between Two Functions

A local linear approximation also lets us compare two functions without computing their difference exactly. If both functions pass through the same point and have the same derivative there, they share the same tangent line, and their difference is small relative to the distance from that point.

Theorem (First-Order Contact): Let $f$ and $g$ be differentiable at an interior point $a$. If $f(a)=g(a)$ and $f'(a)=g'(a)$, then $$ \lim_{h\to0,\ h\ne0}\frac{|f(a+h)-g(a+h)|}{|h|}=0. $$

Proof. Write the remainders for the two local linear approximations as

\begin{aligned} f(a+h)&=f(a)+f'(a)h+R_f(h),\\ g(a+h)&=g(a)+g'(a)h+R_g(h). \end{aligned}

By differentiability, $|R_f(h)|/|h|\to0$ and $|R_g(h)|/|h|\to0$. The hypotheses $f(a)=g(a)$ and $f'(a)=g'(a)$ cancel the constant and linear terms when we subtract the two equations. Thus

|f(a+h)-g(a+h)|=|R_f(h)-R_g(h)| \le |R_f(h)|+|R_g(h)|.

For $h\ne0$, divide by $|h|$. The resulting right-hand side tends to zero, since each of its two terms tends to zero. The nonnegative quotient on the left is bounded above by a quantity tending to zero, so it too tends to zero. $\square$

Worked Example: A Curve and Its Tangent Have First-Order Contact

Let $f(x)=x^2$ and $g(x)=2x-1$, considered at $a=1$. We have $f(1)=1=g(1)$. The difference quotient for $f$ at $1$ is $((1+h)^2-1)/h=2+h$, so $f'(1)=2$; the difference quotient for the affine function $g$ is $2$, so $g'(1)=2$. The theorem applies. Indeed, direct subtraction gives

f(1+h)-g(1+h) =(1+h)^2-\bigl(2(1+h)-1\bigr) =h^2.

Consequently, the absolute difference divided by the distance $|h|$ is $|h|$, which tends to zero. The two functions need not agree at nearby points; rather, their difference becomes small compared with the distance to $1$.

Interpreting the Approximation Carefully

The derivative is a first-order model, not a promise that the graph is a straight line in a neighborhood. For example, the quartic example has a nonzero remainder $6h^2+4h^3+h^4$ whenever $h$ is sufficiently small and nonzero. Its tangent line still gives the correct first-order change because the remainder divided by $|h|$ tends to zero.

The condition $f'(a)=0$ means that the linear part of the change vanishes. In that case differentiability says $f(a+h)-f(a)$ is small relative to $|h|$. It does not say that $f(a+h)=f(a)$, nor does it decide whether nearby values are larger or smaller. For example, $x^2$ at zero has zero derivative but positive values at every nonzero $x$. When the derivative is nonzero, the Local Sign Theorem gives a one-sided comparison with $f(a)$; even then, it does not alone establish monotonicity between arbitrary nearby points.

A final useful check is to keep the scale of the error explicit. To verify a proposed local linear approximation, subtract the proposed line from the function and divide the absolute value of that remainder by $|h|$. The limit must be zero. Merely showing that the remainder tends to zero is not enough: the definition requires it to be small relative to the input change. The First-Order Approximation Characterization supplies this criterion, while the results here show how to use it to compare values and functions.

Check Your Understanding

Use the local linear approximation and its remainder to answer these questions.

If $f'(a)=m$, what expression gives the remainder after subtracting the tangent-line approximation from $f(a+h)$?
For $f(x)=1/(2+x)$ at zero, what are the local linear approximation and the exact remainder?
Why does $f'(a)>0$ imply that sufficiently close points to the right have values greater than $f(a)$, without proving monotonicity on an interval?
If two differentiable functions agree in value and derivative at $a$, what limit does their difference satisfy after division by $|h|$?
Why is it not enough to know only that the remainder tends to zero as $h\to0$?