Derivative of the Identity

The Simplest Nonconstant Function

In “Derivative of a Constant,” the difference quotient was zero because the function’s output did not change when its input changed. The identity function gives a contrasting basic case: every change in the input is reproduced exactly as the same change in the output. Its difference quotient is therefore not zero, but one.

The identity function is defined on an interval, which need not be all of \(\mathbb{R}\). Its derivative is considered at interior points, as in the definition used earlier in this course. At such points there is room to vary the input a small amount in either direction while remaining in the domain.

Definition: Let \(I\subseteq\mathbb{R}\) be an interval. The identity function on \(I\), denoted \(\operatorname{id}_I\), is the function \(\operatorname{id}_I:I\to\mathbb{R}\) defined by \(\operatorname{id}_I(x)=x\) for every \(x\in I\). Its output at each input is that same input.

The identity function is not a constant function: when its input changes, its output changes as well. The two functions can have the same value at a particular point—for instance, both might equal \(4\) there—but their behavior at nearby inputs is different. A constant function keeps its output fixed; the identity function moves by precisely the amount the input moves.

The Difference Quotient Is Exactly One

Theorem (Derivative of the Identity): Let \(I\subseteq\mathbb{R}\) be an interval, and let \(a\) be an interior point of \(I\). The identity function \(\operatorname{id}_I\) is differentiable at \(a\), and \(\operatorname{id}_I'(a)=1\).

Proof. Since \(a\) is an interior point, \(a+h\in I\) for every sufficiently small \(h\). For each such \(h\ne0\), the definition of the identity function gives \(\operatorname{id}_I(a+h)=a+h\) and \(\operatorname{id}_I(a)=a\). Thus

\frac{\operatorname{id}_I(a+h)-\operatorname{id}_I(a)}{h} = \frac{(a+h)-a}{h} = \frac{h}{h} =1.

The difference quotient equals \(1\) for every allowed nonzero \(h\), not merely for values of \(h\) close to zero. Its limit as \(h\to0\) therefore exists and equals \(1\). By the definition of the derivative, \(\operatorname{id}_I'(a)=1\). The point \(a\) was arbitrary, so the result holds at every interior point of \(I\). \(\square\)

The calculation contains no approximation: the quotient is exactly one before taking a limit. The limit is still essential to the definition of the derivative, but here the limit is immediate because the quotient has the same value for every permitted nonzero increment.

Worked Example: An Identity Function on a Bounded Interval

Let \(I=(2,8)\), and define \(f=\operatorname{id}_I\). At \(a=5\), if \(|h|<3\), then \(5+h\in(2,8)\). For every such \(h\ne0\),

\frac{f(5+h)-f(5)}{h} = \frac{(5+h)-5}{h} = \frac{h}{h} =1.

It follows that \(f'(5)=1\). The restriction \(|h|<3\) ensures that the input stays in the domain; it does not change the quotient. The same theorem applies at any other interior point of \((2,8)\), although the permitted size of \(h\) may depend on the point.

Worked Example: A Negative Base Point

Let \(f=\operatorname{id}_{\mathbb{R}}\), and examine the derivative at \(a=-4\). For every \(h\ne0\),

\frac{f(-4+h)-f(-4)}{h} = \frac{(-4+h)-(-4)}{h} = \frac{h}{h} =1.

Consequently, \(f'(-4)=1\). The negative value of the base point introduces no sign change: the terms \(-4\) cancel, leaving \(h\) in the numerator and \(h\) in the denominator.

Exact First-Order Approximation

The First-Order Approximation Characterization from “Definition of the Derivative” describes differentiability in terms of a linear term and a remainder that is small relative to the size of the input change. For the identity function, the remainder is not merely small: it is always zero. This is a stronger statement than the existence of a derivative.

Theorem (Exact First-Order Approximation for the Identity): Let \(I\subseteq\mathbb{R}\), let \(a\in I\), and let \(h\) satisfy \(a+h\in I\). Then \(\operatorname{id}_I(a+h)=\operatorname{id}_I(a)+h\). In particular, at every interior point \(a\), the first-order approximation using the derivative is exact for every allowed increment \(h\).

Proof. By definition, \(\operatorname{id}_I(a+h)=a+h\) and \(\operatorname{id}_I(a)=a\). Therefore

\operatorname{id}_I(a+h) = a+h = \operatorname{id}_I(a)+h.

By the Derivative of the Identity Theorem, \(\operatorname{id}_I'(a)=1\) whenever \(a\) is an interior point. The first-order expression at \(a\) is consequently \(\operatorname{id}_I(a)+\operatorname{id}_I'(a)h=\operatorname{id}_I(a)+h\), which equals \(\operatorname{id}_I(a+h)\) by the displayed identity. Thus the approximation has zero error for every allowed \(h\), including increments that are not especially small. \(\square\)

For a general differentiable function, a nonzero remainder may remain after subtracting the first-order term. The identity function is special in this respect: its graph already is the linear function given by that first-order term. The derivative gives its local linear description, and that description agrees with the function exactly throughout its domain.

Worked Example: Checking the Approximation at a Nonzero Increment

Let \(f=\operatorname{id}_{\mathbb{R}}\), choose \(a=-2\), and take \(h=\tfrac{1}{10}\). The first-order expression at \(a\) is \(f(a)+f'(a)h\). Since \(f(-2)=-2\) and \(f'(-2)=1\), it equals

f(-2)+f'(-2)\left(\frac{1}{10}\right) = -2+\frac{1}{10} = -\frac{19}{10}.

The function value at the displaced input is also

f\left(-2+\frac{1}{10}\right) = f\left(-\frac{19}{10}\right) = -\frac{19}{10}.

Thus the error is exactly zero for this increment. The exact first-order approximation theorem explains why this equality holds not only for \(\tfrac{1}{10}\), but for every increment \(h\) that keeps the input in the domain.

Secant Slopes and the Tangent Line

The derivative can also be understood through secant slopes. For the identity function, any two distinct inputs produce the same slope: the output difference equals the input difference. As a result, the tangent line does not just describe the graph near a point; it is the graph itself.

Theorem (Secant and Tangent Lines to the Identity Graph): Let \(I\subseteq\mathbb{R}\), and let \(u,v\in I\) with \(u\ne v\). The secant slope between \((u,\operatorname{id}_I(u))\) and \((v,\operatorname{id}_I(v))\) is \(1\). At every interior point \(a\in I\), the tangent line to the graph of \(\operatorname{id}_I\) is \(y=x\).

Proof. Since \(\operatorname{id}_I(u)=u\) and \(\operatorname{id}_I(v)=v\), the secant slope is

\frac{\operatorname{id}_I(v)-\operatorname{id}_I(u)}{v-u} = \frac{v-u}{v-u} =1.

The graph point at \(a\) is \((a,a)\), and by the Derivative of the Identity Theorem the tangent slope there is \(1\). The line of slope \(1\) through \((a,a)\) has equation \(y-a=x-a\), which simplifies to \(y=x\). Hence the tangent line is the graph of the identity function itself. \(\square\)

Worked Example: A Secant Through Two Distinct Inputs

For \(f=\operatorname{id}_{\mathbb{R}}\), choose \(u=\tfrac{1}{3}\) and \(v=4\). The corresponding graph points are \((\tfrac{1}{3},\tfrac{1}{3})\) and \((4,4)\). Their secant slope is

\frac{4-\frac{1}{3}}{4-\frac{1}{3}} = \frac{\frac{11}{3}}{\frac{11}{3}} =1.

The secant line through these points is \(y=x\), since both points satisfy that equation. In fact, the same line contains every graph point \((x,f(x))=(x,x)\), so choosing different distinct inputs does not produce a different secant line.

Domain and Interpretation

The identity function depends on its domain, even though its formula is always \(x\). If \(I\) is an interval with endpoints, the derivative theorem applies at its interior points. Under the two-sided derivative definition used here, an endpoint does not have nearby domain points on both sides. A one-sided derivative may be defined separately, but the theorem above makes no claim about one-sided derivatives.

It is also important to distinguish the identity function from a constant function at the level of changes, not just values. For the identity function, an increment \(h\) in the input changes the output by \(h\), and the difference quotient is \(h/h=1\). For a constant function, the output change is zero and the difference quotient is \(0/h=0\). These exact calculations explain the different derivatives without relying on a visual impression of the graphs.

Finally, the value of a derivative is a rate of change, not the value of the function. At a point \(a\), the identity function has value \(a\) and derivative \(1\). For example, at \(a=-4\), its value is \(-4\), while its derivative is \(1\). The derivative does not report the height of the graph; it reports how output changes in relation to input near the point.

Check Your Understanding

Use the definition of the identity function and the difference quotient to answer the following questions.

What is the identity function on an interval \(I\), and how does it differ from a constant function?
At an interior point \(a\), simplify the difference quotient for \(\operatorname{id}_I\) and explain why its limit exists.
Why is the derivative of the identity function one even when the base point \(a\) is negative?
In what sense is the first-order approximation of the identity function exact?
What is the secant slope between two distinct points on the graph of the identity function, and what does this imply about its tangent line?
Why does the stated theorem concern interior points rather than endpoints of the domain?

What You'll Learn