A method
for approximating the value of a function near
a known value. The method uses the tangent
line at the known value of the function
to approximate the function's graph.
In this method Δx and
Δy represent
the changes in x and y for the function, and dx and dy represent
the changes in x and y for the tangent line.
Example:
Approximate \(\sqrt {10} \) by
differentials.
Solution:
\(\sqrt {10} \) is
near \(\sqrt 9 \),
so we will use \(f\left( x \right) = \sqrt x \) with x =
9 and Δx =
1. Note that \(f'\left( x \right) = \frac{1}{{2\sqrt x }}\).
\(\eqalign{
\sqrt {10} &= f\left( {x + \Delta x} \right)\\ &\approx f\left( x \right) + f'\left( x \right)\Delta x\\ &= \sqrt x + \frac{1}{{2\sqrt x }}\Delta x\\ &= \sqrt 9 + \frac{1}{{2\sqrt 9 }}\left( 1 \right)\\ &= 3\frac{1}{6}
}\)
.
Thus we see that \(\sqrt {10} \approx 3\frac{1}{6} = 3.166\bar 6\). This is very close to the correct value of \(\sqrt {10} \approx 3.1623\).
f(x) = The known (exact) value of the function at the base point x
f′(x) = The derivative of the function evaluated at the base point x
Δx = The small change in x from the base point to the target point (also written as dx)
f(x+Δx) = The function value you want to approximate
Worked Example
Problem: Use differentials to approximate the cube root of 27.5, i.e., estimate ∛(27.5).
Step 1: Choose a nearby value where the function is easy to compute. Since 27.5 is close to 27, let f(x) = x^(1/3), x = 27, and Δx = 0.5.
f(x)=x1/3,x=27,Δx=0.5
Step 2: Find the derivative and evaluate it at x = 27.
f′(x)=31x−2/3⟹f′(27)=31(27)−2/3=3⋅91=271
Step 3: Apply the differential approximation formula.
f(27.5)≈f(27)+f′(27)⋅Δx=3+271(0.5)
Step 4: Compute the numerical result.
327.5≈3+270.5=3+0.01852≈3.0185
Step 5: Compare with the actual value: ∛(27.5) ≈ 3.01846. The approximation is accurate to four decimal places.
Answer: ∛(27.5) ≈ 3.0185
Another Example
This example involves a trigonometric function instead of a root, and requires converting degrees to radians for Δx — a common source of error.
Problem: Use differentials to approximate sin(31°).
Step 1: Pick a nearby angle with a known sine value. Since 31° is close to 30°, set f(x) = sin(x) with x = 30° = π/6. The change is Δx = 1° = π/180 radians. You must convert to radians because the derivative formulas for trig functions require radian measure.
x=6π,Δx=180π
Step 2: Find the derivative and evaluate it at x = π/6.
Step 5: The actual value is sin(31°) ≈ 0.5150. The estimate is extremely close.
Answer: sin(31°) ≈ 0.5151
Frequently Asked Questions
What is the difference between Δy and dy in differential approximation?
Δy is the actual change in the function's value: Δy = f(x + Δx) − f(x). The differential dy = f'(x) dx is the change along the tangent line over the same horizontal distance. The approximation works because dy ≈ Δy when Δx is small. The gap between them is the approximation error.
When should you use differentials to approximate a value?
Use this method when you need a quick estimate of a function near a point where the exact value is easy to compute. It is especially useful for roots, powers, and trig values that are close to 'nice' numbers. If Δx is large relative to x, the tangent-line estimate becomes less accurate, so the method works best for small changes.
Is approximation by differentials the same as linearization?
Yes, they are essentially the same idea. Linearization writes the tangent-line approximation as L(x) = f(a) + f'(a)(x − a), and approximation by differentials writes it as f(a + Δx) ≈ f(a) + f'(a)Δx. Both use the first-order Taylor expansion about the point a.
Approximation by Differentials (Linear Approximation) vs. Taylor Polynomial (Higher-Order Approximation)
Approximation by Differentials (Linear Approximation)
Taylor Polynomial (Higher-Order Approximation)
Degree
Uses a first-degree (linear) polynomial — the tangent line
Can use polynomials of any degree n
Formula
f(a) + f'(a) Δx
f(a) + f'(a) Δx + f''(a)(Δx)²/2! + ⋯
Accuracy
Good for very small Δx; error grows quickly as Δx increases
Higher-degree terms improve accuracy for larger Δx
Complexity
Requires only f and f' at one point
Requires higher-order derivatives f'', f''', etc.
When to use
Quick estimates near a known value
When greater precision is needed or Δx is not very small
Why It Matters
Approximation by differentials appears in AP Calculus AB/BC free-response and multiple-choice questions regularly. Engineers and scientists use it to estimate how small measurement errors propagate through formulas — for instance, how a small error in a radius affects the computed volume of a sphere. Understanding this technique also lays the groundwork for Taylor series, which generalize the idea to higher-order approximations.
Common Mistakes
Mistake: Choosing a base point x where f(x) is not easy to compute exactly.
Correction: Always pick a base point where you know the exact function value (e.g., perfect squares for √x, multiples of 30° or 45° for trig). The whole method depends on having one exact value to start from.
Mistake: Forgetting to convert degrees to radians when approximating trigonometric functions.
Correction: The derivative formulas d/dx[sin x] = cos x and d/dx[cos x] = −sin x are valid only in radians. If Δx is given in degrees, convert it to radians (multiply by π/180) before plugging into the formula.
Related Terms
Differential — The dy = f'(x)dx used in the approximation
Tangent Line — The linear model underlying the approximation
Linearization — Equivalent name for the same technique
Derivative — Provides the slope used in the formula
