Estimating Derivatives of a Function at a Point
Introduced| English | Meaning |
|---|---|
| estimate/ˈestɪmət/ | An approximation — say so, and use the closest data you have. |
| difference quotient/ˈdɪfrəns ˈkwəʊʃənt/ | The change in a function's output divided by the corresponding change in its input over an interval. |
Measuring a slope you can't compute exactly
- Sometimes you have a graph or a table, not a formula — but you still want $f'$ at a point.
- You can estimate the derivative from what you have.
- From a graph: measure the steepness of the tangent line by eye.
- From a table: use a difference quotient over the closest available inputs.
Estimate the slope by eye
y = x²
Move the point and watch the tangent — its steepness is the derivative you would estimate from a graph.
From a graph: tangent steepness
- Draw (or imagine) the tangent line at the point, then read its slope as rise ÷ run.
- Pick two clear points on the tangent, not on the curve, and compute $\tfrac{\Delta y}{\Delta x}$.
- Steeper tangent → larger $|f'|$; a flat tangent → $f'\approx0$.
- Uphill tangent → positive derivative; downhill → negative.
The instantaneous rate of change is the gradient of the tangent at a point
To estimate $f'(a)$ from a graph, you should measure the slope of the...
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The derivative is the tangent slope, read from points on the tangent line.
From a table: nearest difference quotient
- With table values, approximate $f'(a)$ by the difference quotient over the closest points:
- A symmetric estimate is best: $f'(a)\approx\dfrac{f(a+h)-f(a-h)}{2h}$, using one point on each side.
- Or use a one-sided quotient $\dfrac{f(a+h)-f(a)}{h}$ if only one neighbor is available.
- Smaller $h$ (closer inputs) generally gives a better estimate 估计值.
A table gives $f(1.9)=4.2$ and $f(2.1)=5.8$. Use the symmetric difference quotient to estimate $f'(2)$.
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$\dfrac{5.8-4.2}{2(0.1)}=\dfrac{1.6}{0.2}=8$.
Using input values closer to the point generally improves a table-based derivative estimate.
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A smaller interval better approximates the tangent slope.
A derivative read from a table or graph is an ____, not an exact value.
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Only algebra on a formula gives the exact derivative.
From $f(3)=10$ and $f(3.2)=11.4$, estimate $f'(3)$ with a one-sided difference quotient.
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$\dfrac{11.4-10}{3.2-3}=\dfrac{1.4}{0.2}=7$.
Reading sign and size
- The sign of the estimate says which way $f$ is heading: $+$ rising, $-$ falling.
- The size says how fast: a derivative of $12$ means $f$ changes about $12$ units per unit of $x$ there.
- Always attach units and interpret in context (e.g. "the temperature is rising about $2^\circ$C per minute").
- An estimate is an approximation — say so, and use the closest data you have.
If an estimated derivative at a point is negative, the function there is...
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A negative rate means the function is falling there.
When estimating from a graph, read the slope of the tangent line, not of a nearby secant, and take your two points off the tangent, not off the curve. From a table, don't reuse the point $a$ itself as both inputs — you need two different inputs bracketing (or approaching) $a$.
A table gives $f(1.9)=4.2$, $f(2)=5.0$, $f(2.1)=5.9$. Estimate $f'(2)$.
- Symmetric difference quotient: $f'(2)\approx\dfrac{f(2.1)-f(1.9)}{2(0.1)}=\dfrac{5.9-4.2}{0.2}=\dfrac{1.7}{0.2}=8.5$.
- (A one-sided estimate $\dfrac{5.9-5.0}{0.1}=9$ is close but less balanced.)
- So $f'(2)\approx8.5$ — $f$ is rising at about $8.5$ units per unit of $x$.
To estimate a derivative: from a graph, read the slope of the tangent line (rise ÷ run using points on the tangent); from a table, use a difference quotient over the closest inputs — the symmetric quotient $\frac{f(a+h)-f(a-h)}{2h}$ is most accurate. The sign gives direction, the size gives rate.