From a Difference Quotient to a Constant Derivative
The identity function returns its input unchanged. Its rule looks simple, but it is useful for seeing directly how a derivative measures local change. Applying the difference quotient requires only substitution and simplification: the change in output is exactly the change in input.
Recall that the derivative at an input is defined as the limit of the difference quotient, if that limit exists. The quotient measures the average rate of change over an input interval of length \(h\); taking the limit as \(h\) approaches zero gives the derivative at the starting input. For the identity function, the quotient is the same no matter where the interval starts or how small the nonzero interval is.
Why the Derivative Is One
Fix any real input \(a\). When the input changes from \(a\) to \(a+h\), the identity function’s output changes from \(a\) to \(a+h\). Thus the output change is \(h\), exactly equal to the input change. Since the difference quotient divides the output change by the input change, its value is one for every nonzero \(h\).
This equality holds for positive and negative \(h\). Therefore both sides of \(a\) give the same quotient, and the limit as \(h\) approaches zero exists. The following derivation gives the result at an arbitrary real input, rather than just checking a particular example.
For an arbitrary \(a\in\mathbb{R}\), form the difference quotient. The restriction \(h\ne0\) is required because it appears in the denominator.
Since \(f(a+h)=a+h\) and \(f(a)=a\), the quotient becomes \(\frac{(a+h)-a}{h}\).
The numerator is \(h\), so for every \(h\ne0\), the quotient is \(\frac{h}{h}=1\).
A quantity that equals one for every nonzero \(h\) near zero has limit one. Hence \(f'(a)=1\).
The cancellation is valid because the difference quotient is considered only for \(h\ne0\). We do not substitute \(h=0\) into the quotient. Instead, we simplify it for nonzero \(h\), then find its limit. Because \(a\) was arbitrary, the calculation proves the result at every real input.
Worked Example: Derivative at a Specific Input
Worked Example: Derivative at a Specific Input
Find the derivative of \(f(x)=x\) at \(a=5\) directly from the difference quotient.
Begin with the definition and substitute \(a=5\):
Using \(f(x)=x\), we have \(f(5+h)=5+h\) and \(f(5)=5\). Therefore, for \(h\ne0\),
The quotient is one for every nonzero \(h\), so its limit as \(h\) approaches zero is one. Thus \(f'(5)=1\). Checking against the general result, the derivative function has value \(1\) at \(x=5\), as required.
Worked Example: Derivative at an Arbitrary Real Input
Worked Example: Derivative at an Arbitrary Real Input
Let \(a=-3\). Compute \(f'(-3)\) from the definition, including the values of the function in the quotient.
The starting input is \(-3\), and the nearby input is \(-3+h\). Since \(f(x)=x\),
Substitute these values into the quotient:
Taking the limit gives \(f'(-3)=1\). The negative starting input changes the function values but does not change the amount by which the output changes when the input changes by \(h\). Checking the result in the derivative function \(f'(x)=1\) also gives \(f'(-3)=1\).
Worked Example: Secant Slopes and the Graph
Worked Example: Secant Slopes and the Graph
Consider the two points on the graph \(y=f(x)=x\) with inputs \(x_1=-2\) and \(x_2=3\). Find the slope of the secant line between them, then compare it with the derivative at \(x_1\).
The function values are \(f(-2)=-2\) and \(f(3)=3\). The secant slope is the change in output divided by the change in input:
The derivative at the first point is \(f'(-2)=1\), by the difference-quotient calculation. The secant slope and derivative therefore agree. In fact, the graph is the straight line \(y=x\), so the secant between any two distinct points on it lies along that same line. The line’s slope is one: for each increase of one unit in \(x\), \(y\) increases by one unit.
As a check, the points \((-2,-2)\) and \((3,3)\) both satisfy \(y=x\), and their rise and run are each \(5\). Their ratio is \(5/5=1\), consistent with both the secant calculation and the derivative.
Interpreting the Constant Slope
The graph of \(f(x)=x\) is a straight line through the origin. A line’s slope describes how much its vertical coordinate changes for a given change in its horizontal coordinate. Here those changes are always equal, so their ratio is always one. This is true for a finite interval between distinct inputs, as in the example, and the derivative calculation shows that the limiting slope at each point is also one.
In geometric language, the derivative at a point gives the slope of the graph’s tangent there. For this function, the graph itself is already a straight line, so the tangent at any point has the same slope as the line: one. The derivative does not vary with the input. That constant value expresses the uniform rate of change of the identity function.
When the input and output represent quantities with units, the derivative’s units are output units per input unit. For example, if both axes use the same unit, a slope of one means one output unit of change for each input unit of change. The numerical derivative is one; the interpretation of its units depends on what the axes represent.
A Common Mistake: Confusing Function Values with Derivative Values
The function value and derivative value answer different questions. The value \(f(a)\) is the output at input \(a\). The value \(f'(a)\) is the local rate of change there. For the identity function, they happen to be related by \(f(a)=a\) and \(f'(a)=1\); they are not generally the same number.
For instance, a mistaken substitution might treat \(f(5+h)\) as \(5\), the value at the starting input, rather than as the value at the changed input. That produces the incorrect calculation
The correction is to apply the function rule to the entire new input: \(f(5+h)=5+h\), not \(5\). Thus the correct quotient is
The wrong calculation ignores the output change caused by changing the input. Keeping the two function values distinct until after substitution prevents this error.
Why This Result Is a Useful Starting Point
The identity function provides a direct example of the central operation in differentiation: compare nearby function values, divide by the input change, and take a limit. Its difference quotient simplifies completely because output and input change by the same amount. The result gives a basic reference point for understanding derivative values and graph slopes.
The conclusion is specific: for \(f(x)=x\), the derivative is one everywhere on \(\mathbb{R}\). It does not say that every function has derivative one, nor that a derivative always equals the function’s value. The difference quotient must be formed for the function under consideration, and its limit must exist. Here the quotient is exactly one for every allowed \(h\), which establishes both existence and value without any additional assumptions.
Check Your Understanding
Answer these questions using the definition and the graph interpretation.
- For \(f(x)=x\), write the difference quotient at an arbitrary input \(a\), and simplify it for \(h\ne0\).
- Why is it valid to cancel \(h\) in the difference quotient but not to substitute \(h=0\) before simplifying?
- What is \(f'(-7)\), and what feature of the difference quotient makes its value independent of the starting input?
- Find the secant slope between the points on \(y=x\) with inputs \(x=-1\) and \(x=4\). How does it relate to the derivative?
- For \(f(x)=x\), what is \(f(5)\), and what is \(f'(5)\)? Explain why these values represent different things.
