From an Instantaneous Rate to a Local Model
The derivative at a point describes how a function changes at that point. It can also help estimate nearby function values. The connection comes from looking closely at the definition of the derivative: a difference quotient compares a change in function output with a change in input.
Fix a point \(a\), and let \(h\) be a nonzero change in the input. The new input is \(a+h\), so the corresponding change in the function value is \(f(a+h)-f(a)\). If \(f\) is differentiable at \(a\), then the average rate of change over this small input change approaches \(f'(a)\) as \(h\) approaches zero.
The key step is to use the quotient’s tendency toward \(f'(a)\) to describe the function change itself. Multiplying the quotient by \(h\) gives the exact change \(f(a+h)-f(a)\). When \(h\) is small, the quotient is close to \(f'(a)\), so the change in the function is close to \(f'(a)h\). This leads to the approximation
The approximation is not an arbitrary formula: it follows from the derivative limit. The next sections make that reasoning precise and show how the resulting expression is the tangent line at the point.
Deriving the Approximation from the Limit
Assume \(f\) is differentiable at \(a\). Differentiability means that the difference quotient approaches the finite number \(f'(a)\). For nonzero \(h\), write that quotient as \(f'(a)\) plus the amount by which it differs from \(f'(a)\):
Multiplying both sides by \(h\) and rearranging gives an exact identity:
The expression in parentheses approaches zero as \(h\) approaches zero, by the definition of \(f'(a)\). Thus the last term is \(h\) multiplied by a quantity that approaches zero. It is a remainder that becomes small compared with \(h\). In particular, the difference between the function value and the linear expression, divided by \(h\), approaches zero.
The limit condition on \(r(h)\) is the precise meaning of a first-order approximation. It says that the error becomes negligible compared with the size of the input change. It does not say that the error is always zero, or that it has a particular numerical bound for every chosen \(h\).
Why the Linear Expression Is a Tangent Line
As a function of the new input \(x=a+h\), the expression \(f(a)+f'(a)h\) can be rewritten using \(h=x-a\). It becomes
This is a line. At \(x=a\), its value is \(L(a)=f(a)\), so it passes through the point \((a,f(a))\). Its slope is \(f'(a)\), the derivative of the function at that point. Therefore it is the tangent line to the graph of \(f\) at \((a,f(a))\). The approximation is the claim that near \(a\), this tangent line models the function’s values well.
Changing \(a\) to \(a+h\) changes the function value by \(f(a+h)-f(a)\).
The average rate of change \(\frac{f(a+h)-f(a)}{h}\) approaches \(f'(a)\) as \(h\) approaches zero.
The function change is approximately \(f'(a)h\), so the nearby value is approximately \(f(a)+f'(a)h\).
With \(x=a+h\), this expression is the line through \((a,f(a))\) with slope \(f'(a)\).
Worked Examples from the Difference Quotient
Worked Example: Deriving the Approximation for a Square
Let \(f(x)=x^2\) and use the base point \(a=3\). First compute the derivative from its definition:
The function value at the base point is \(f(3)=9\). The tangent-line expression for an input \(3+h\) is therefore \(9+6h\). For \(h=-\frac{1}{10}\), the estimate is
Check against the exact value: \(3-\frac{1}{10}=\frac{29}{10}\), so
The estimate is \(8.4\), while the exact value is \(8.41\). Their difference is \(8.41-8.4=0.01\). Directly, the exact expansion is \(f(3+h)=9+6h+h^2\), so the difference from the tangent-line expression is \(h^2\); at \(h=-\frac{1}{10}\), that is \(\frac{1}{100}=0.01\). The check agrees.
Worked Example: Deriving the Approximation for a Cube
Let \(f(x)=x^3\) and \(a=1\). Apply the derivative definition and expand the numerator:
Since \(f(1)=1\), the linear expression at \(1+h\) is \(1+3h\). For \(h=\frac{1}{10}\), it gives
Check this estimate by calculating the exact value:
The exact value exceeds the estimate by \(1.331-1.3=0.031\). From the expansion, the remainder after \(1+3h\) is \(3h^2+h^3\). Substituting \(h=\frac{1}{10}\) gives \(3(\frac{1}{100})+\frac{1}{1000}=\frac{30}{1000}+\frac{1}{1000}=\frac{31}{1000}=0.031\), which verifies the difference.
Worked Example: Deriving the Approximation for a Reciprocal
Let \(f(x)=\frac{1}{x}\) and use \(a=2\). For \(h\) near zero, \(2+h\ne 0\), so the difference quotient is defined:
Because \(f(2)=\frac{1}{2}\), the tangent-line expression at \(2+h\) is \(\frac{1}{2}-\frac{h}{4}\). Take \(h=\frac{1}{10}\):
The exact value is \(f(\frac{21}{10})=\frac{10}{21}\). The exact value minus the estimate is
This also checks the estimate directly: \(\frac{19}{40}+\frac{1}{840}=\frac{399}{840}+\frac{1}{840}=\frac{400}{840}=\frac{10}{21}\). The tangent-line value is close to, but not equal to, the function value.
What the Approximation Does—and Does Not—Say
The derivative definition is a limit as \(h\) approaches zero. It justifies a local approximation: for small changes in input, the tangent line gives a close model of the function. It does not, by itself, say how close the estimate is for a particular nonzero \(h\). The size of the actual error depends on the function and on the chosen change.
A frequent mistake is to replace the approximation symbol with an equality for every \(h\). For example, for \(f(x)=x^2\) at \(a=3\), the tangent-line expression is \(9+6h\). A mistaken calculation might assert
But expanding the left side gives
The omitted term \(h^2\) is generally not zero. The correct conclusion is that \(9+6h\) approximates \((3+h)^2\) when \(h\) is close to zero; the exact difference is \(h^2\). The tangent line and the graph meet at the base point and have the same slope there, but they need not coincide elsewhere.
Check Your Understanding
Use the difference quotient and the meaning of the remainder to answer these questions.
- For a differentiable function at \(a\), what limit does the difference quotient approach as \(h\) approaches zero?
- Explain why multiplying the difference quotient by \(h\) suggests that the function change is approximately \(f'(a)h\).
- Why does \(f(a)+f'(a)h\) describe a tangent line when the input is \(a+h\)?
- For \(f(x)=x^2\) at \(a=3\), what exact term is omitted from the tangent-line expression?
- Does differentiability alone make the tangent-line estimate exact for every nearby input? Explain.