Why small setup errors matter
A linear estimate uses a tangent line at a chosen base point to approximate a function nearby. The calculation is often short, but its result depends on matching the derivative to the correct base point, using the signed change in the input, and keeping units consistent. A separate source of confusion is treating the differential as if it were the exact change in the function.
These are not merely notation issues. A sign reversal can turn an estimate of a decrease into an increase; a derivative at the wrong point changes the tangent line; and mixing centimetres with metres can change a numerical result by a factor of one hundred or more.
Keep the base point fixed
The base point \(a\) is the input where the tangent line is anchored. Both the function value \(f(a)\) and the derivative \(f'(a)\) in the linearization must be evaluated there. The target input \(x\) determines the increment \(x-a\), but it does not replace \(a\) in the derivative.
Worked Example: Using the derivative at the target by mistake
Estimate \(f(3.2)\) for \(f(x)=x^2\), using the base point \(a=3\).
First calculate the input change and the derivative at the base point:
The linearization gives:
A common error is to use \(f'(3.2)=6.4\), the derivative at the target, while still using the base-point value \(f(3)=9\). That calculation gives \(9+6.4(0.2)=9+1.28=10.28\), which is not the linearization at \(3\).
Check the correct estimate against the function value: \(f(3.2)=(3.2)^2=10.24\). The estimate \(10.2\) is the value from the tangent line at \(3\), as the formula specifies. In particular, it uses \(f'(3)=6\), not \(f'(3.2)\).
Preserve the sign of the increment
The increment is defined by the target input minus the base point: \(\Delta x=x-a\). It is negative when the target is below the base point and positive when the target is above it. Since the linear change is \(f'(a)\Delta x\), reversing the subtraction reverses the sign of the estimated change.
Worked Example: Estimating a logarithm below the base point
Use the linearization of \(f(x)=\ln x\) at \(a=4\) to estimate \(\ln(3.9)\).
The derivative is \(f'(x)=1/x\), so \(f'(4)=1/4\). The target is below the base point, giving the signed increment:
Therefore, the estimate is:
Using \(\ln 4\approx1.386294\), rounded to six decimal places, gives \(L_4(3.9)\approx1.361294\). If the increment is mistakenly reversed to \(4-3.9=0.1\), the calculation becomes \(\ln 4+\frac14(0.1)=\ln 4+0.025\approx1.411294\), which moves in the wrong direction.
Check the sign by substituting the target into the linearization formula: \(x-a=3.9-4=-0.1\), and \(f'(4)(x-a)=(1/4)(-0.1)=-0.025\). The estimate must subtract \(0.025\) from \(\ln 4\), as calculated.
Carry units through the calculation
Units are part of the quantities, not labels to add after the arithmetic. If \(y=f(x)\), the derivative \(f'(a)\) has units of output per unit of input. Thus \(dy=f'(a)\,dx\) has units of output. The base value, derivative, and input change must use compatible units.
Worked Example: Mixing centimetres and metres
The area of a circle is \(A(r)=\pi r^2\). Estimate the area change when a radius of \(10\) centimetres increases by \(0.2\) centimetres.
Differentiate with respect to the radius: \(A'(r)=2\pi r\). In centimetres, the base radius is \(r=10\text{ cm}\) and the radius change is \(dr=0.2\text{ cm}\). Therefore:
The units check: area per length, multiplied by a length, gives area. The same calculation in metres uses \(r=0.10\text{ m}\) and \(dr=0.002\text{ m}\):
Since \(1\text{ m}^2=10{,}000\text{ cm}^2\), this is \(0.0004\pi(10{,}000)=4\pi\text{ cm}^2\), matching the first calculation.
A unit-mixing error would use the numerical radius \(10\) as though it were in metres, but keep \(dr=0.002\text{ m}\). It would give \(2\pi(10)(0.002)=0.04\pi\text{ m}^2\), one hundred times the correct \(0.0004\pi\text{ m}^2\). The error is that \(10\) represents centimetres in the original measurement, not metres. Converting both radius quantities first verifies the correct differential.
Distinguish \(dy\) from \(\Delta y\)
The differential \(dy=f'(a)\,dx\) is the change predicted by the tangent-line model at \(a\). The actual change from \(a\) to \(a+\Delta x\) is \(\Delta y=f(a+\Delta x)-f(a)\). Taking \(dx=\Delta x\) makes \(dy\) a linear estimate of \(\Delta y\), but the two quantities need not be equal.
Worked Example: A differential is not the exact change
Let \(f(x)=x^2\), with base point \(a=2\) and input change \(\Delta x=0.5\). Find the differential estimate and the actual change.
Since \(f'(x)=2x\), the derivative at the base point is \(f'(2)=4\). Set \(dx=\Delta x=0.5\). Then:
The actual change uses the function values at the two inputs:
Thus \(dy=2\), whereas \(\Delta y=2.25\). To check the distinction exactly, expand the change for this function:
The differential retains the linear term \(2a\Delta x=2\); the actual change also includes the squared term \(0.25\). Calling \(dy\) the exact change would therefore be incorrect in this example.
Why the base-point formula works
The derivative at \(a\) describes the limiting ratio of the function change to the input change near \(a\). Specifically, differentiability means that as \(h\) approaches \(0\), the difference quotient approaches \(f'(a)\). This is the reason the derivative in the linearization must be taken at the base point.
This argument also clarifies the role of a differential: \(f'(a)\,dx\) is the linear part of the predicted output change. The actual change can include additional terms, so a differential and a finite change should not be identified without justification.
A compact error check
Before accepting a linear estimate, inspect the setup rather than only the final arithmetic. In particular, confirm the base point, the signed increment, and the units before multiplying. Then decide whether the question asks for the differential or for the actual function change.
Write \(a\) explicitly. Evaluate both \(f(a)\) and \(f'(a)\) at this same input.
Use \(\Delta x=x-a\), not its absolute value, unless the question specifically asks for a magnitude.
Convert measurements to compatible units before using the derivative, and check that the result has the units of the output.
Use \(dy=f'(a)\,dx\) for the differential estimate; use \(\Delta y=f(a+\Delta x)-f(a)\) for the actual change.
Each check catches a different mistake. Correcting the sign does not fix a derivative taken at the wrong point; converting units does not make a differential equal to an exact change. Verify each part of the setup independently.
Check Your Understanding
For each question, identify the relevant setup and explain how you would check it.
- To estimate \(f(5.1)\) using a base point \(a=5\), which input must be used in \(f'(a)\), and what is the signed increment?
- For a target \(x=2.8\) and base point \(a=3\), should the linear estimate add or subtract \(f'(3)(0.2)\)? Explain using \(\Delta x=x-a\).
- A radius is given as \(50\) centimetres and a change as \(0.01\) metres. What must be done before multiplying them in a differential formula?
- In general, how are \(dy=f'(a)\,dx\) and \(\Delta y=f(a+\Delta x)-f(a)\) defined, and why need they not agree?
- A calculation uses \(f'(x)\) evaluated at the target while retaining \(f(a)\) and \(x-a\). Which part of the linearization setup is inconsistent?
