Why Small Setup Errors Matter
A tangent-line estimate is built from a few pieces of information: a base point, the function value there, the derivative there, and the change in input. The calculation is short, so it is easy to overlook one piece or use a correct value in the wrong role.
The most common errors are not difficult calculus errors. They are setup errors: reporting the slope as though it were a function value, using derivative information from a different base point, treating a differential as an exact finite change, or forgetting to multiply by the input change.
In this tutorial, \(a\) is the base point and \(x\) is the input whose function value we want to estimate. The input change from \(a\) to \(x\) is \(x-a\).
Keep the Roles of the Quantities Straight
Each part of the formula has a distinct meaning. The value \(f(a)\) is the function’s height at the base point. The value \(f'(a)\) is the tangent line’s slope there. The factor \(x-a\) is how far the input moves from the base point. Their product, \(f'(a)(x-a)\), is the tangent line’s predicted change in output.
This is why the derivative alone is not the estimate. A slope describes how the output changes per unit of input near a point; it does not give the output’s value there. Even when the derivative is the right slope, the estimate must also include the starting value and the input change.
Worked Example: Do Not Use the Derivative as the Estimate
Use the tangent line to estimate \(f(3.1)\) for \(f(x)=x^2\), using the base point \(a=3\).
First find the function value and derivative at the base point:
The input change is \(3.1-3=0.1\). Therefore,
The estimate is \(9.6\). The incorrect answer \(6\) is just the tangent slope at \(x=3\); it is not an estimate of \(f(3.1)\). To check, the exact value is \(f(3.1)=(3.1)^2=9.61\), so the estimate \(9.6\) is close to the exact value.
Use One Consistent Base Point
The base point determines both the point on the curve and the slope of the tangent line. In \(L(x)=f(a)+f'(a)(x-a)\), the same \(a\) must be used in all three places: \(f(a)\), \(f'(a)\), and \(x-a\). Mixing values from different base points does not describe the tangent line at either point.
A useful check is to set \(x=a\). The input change is then zero, so the estimate must return the function’s value at the base point: \(L(a)=f(a)+f'(a)(a-a)=f(a)\). If a proposed expression fails this check, its pieces have likely been combined incorrectly.
Worked Example: Correct a Base-Point Mix-Up
Estimate \(f(2.1)\) for \(f(x)=x^2\) using the base point \(a=2\). Then examine an estimate formed by using the target \(2.1\) as the base point while still adding a step of \(0.1\).
At the correct base point,
The target is \(2.1\), so the input change from \(2\) is \(2.1-2=0.1\). The correct estimate is
Now consider the inconsistent calculation \(f(2.1)+f'(2.1)(0.1)\). It gives
This is not the tangent-line estimate based at \(2\): it uses the target’s function value and slope, then adds another nonzero input change. If \(2.1\) itself is the base point, the input change to estimate \(f(2.1)\) is \(2.1-2.1=0\), and the result is simply \(f(2.1)=4.41\). The correct estimate based at \(2\), \(4.4\), is close to the exact value \(4.41\).
Distinguish a Differential from an Exact Change
For a small input change \(\Delta x\), the corresponding exact change in the function is \(\Delta y=f(a+\Delta x)-f(a)\). The differential \(dy\), when evaluated at \(x=a\) with \(dx=\Delta x\), is \(dy=f'(a)\,dx=f'(a)\Delta x\). It is the change predicted by the tangent line, not generally the exact change in the function.
The reason the differential gives a useful estimate is that differentiability means the function’s change is well approximated by its tangent-line change when the input change is small. But “well approximated” does not mean “identical.” Confusing these quantities hides the approximation error.
Worked Example: Compare \(dy\) with \(\Delta y\)
For \(f(x)=x^2\), start at \(a=3\) and take \(dx=\Delta x=0.2\). Find the differential and the exact change.
The derivative at the base point is \(f'(3)=2(3)=6\). Thus,
The exact change is calculated from the function values:
Therefore \(dy=1.2\), while \(\Delta y=1.24\); their difference is \(1.24-1.2=0.04\). The corresponding tangent-line estimate of the new function value is \(f(3)+dy=9+1.2=10.2\). The exact value is \(f(3.2)=10.24\), which confirms that the differential supplies an estimate rather than the exact change.
Why the Tangent-Line Formula Has This Form
A tangent line at \(x=a\) must pass through the point \((a,f(a))\) and have slope \(f'(a)\). A line with slope \(m\) passing through \((a,f(a))\) has point-slope equation \(y-f(a)=m(x-a)\). Substituting \(m=f'(a)\) gives the tangent-line equation.
Replacing the tangent-line output \(y\) with \(L(x)\) gives \(L(x)=f(a)+f'(a)(x-a)\). The term \(x-a\) is essential: it measures the movement from the point where the tangent line is anchored. The formula also shows directly that \(L(a)=f(a)\), since the slope contribution is zero at the base point.
For differentials, write the input change as \(dx\). The tangent-line change in output is its slope multiplied by that input change, so at the base point \(a\), \(dy=f'(a)\,dx\). If \(dx=0\), then \(dy=0\), as expected: with no change in input, the tangent line predicts no change in output.
Do Not Omit the Input Change
A common near-miss is to calculate \(f'(a)\) correctly but forget to multiply it by \(x-a\), or by \(dx\) when using differentials. This error can be especially misleading when the input change is small: it may look like a minor arithmetic omission, but it changes the meaning of the calculation.
Worked Example: Restore the Missing Input Change
Estimate \(\sqrt{4.04}\) by using \(f(x)=\sqrt{x}\) and the base point \(a=4\).
At the base point,
The input change is \(4.04-4=0.04\). The correct estimate is
If the input change is omitted, the incorrect calculation gives \(f(4)+f'(4)=2+\frac14=2.25\). That adds the slope as if the input had changed by one unit, even though the actual input change is only \(0.04\). Squaring the estimate provides a check: \((2.01)^2=4.0401\), which is close to \(4.04\). The incorrect value \(2.25\) gives \((2.25)^2=5.0625\), far from the target input. The exact value is \(\sqrt{4.04}\approx2.009975\), rounded to six decimal places, so \(2.01\) is a reasonable estimate.
A Short Checklist for Diagnosing an Estimate
When a tangent-line estimate or differential calculation looks suspicious, check its structure before redoing the arithmetic. Identify the base point explicitly, write the input change, and then check what each quantity represents.
Record \(a\), the point where the function value and derivative are evaluated.
For a target \(x\), calculate \(x-a\). For a differential estimate, identify \(dx\).
Use \(f(a)+f'(a)(x-a)\), or use \(dy=f'(a)\,dx\) to estimate the change.
A slope is not a function value, and a differential is not generally an exact finite change.
Common Mistakes at a Glance
| Mistake | Why it fails | Correction |
|---|---|---|
| Using \(f'(a)\) as the estimate for \(f(x)\) | \(f'(a)\) is a slope, not an output value. | Use \(f(a)+f'(a)(x-a)\). |
| Mixing base points | The function value, slope, and input change no longer describe one tangent line. | Use the same \(a\) in every part of the formula. |
| Setting \(dy=\Delta y\) without checking | The differential is the tangent-line change; \(\Delta y\) is the exact change. | Use \(dy\) as an approximation unless the function’s behavior makes the changes equal. |
| Leaving out \(x-a\) or \(dx\) | The slope contribution must be scaled by the size of the input change. | Multiply the derivative by the input change before adding it to the starting value. |
These checks matter whenever a local linear model is used. A correctly computed derivative cannot compensate for a missing input change or an inconsistent base point. Keeping the roles of the quantities separate makes the estimate easier to construct and easier to verify.
Check Your Understanding
For each question, pay attention to what each quantity represents before calculating.
- For \(f(x)=x^2\), estimate \(f(4.1)\) using \(a=4\). What is the base-point function value, what is the slope, and what is the input change?
- Explain why \(f'(a)\) alone cannot generally be an estimate of \(f(x)\).
- For a finite input change \(dx\), state the difference between \(dy\) and \(\Delta y\).
- A student uses \(f(a)+f'(a)\) to estimate \(f(a+0.02)\). What factor is missing, and how should the expression be corrected?
- Why should \(L(a)=f(a)\) for a tangent-line approximation based at \(a\)?
