Linearization and Differentials
Write the linear approximation L(x) = f(a) + f'(a)(x−a) and use it to estimate values. Define dy and dx and use differentials to approximate changes. Assess accuracy of approximations using the second derivative.
A lesson of AP Calculus AB, in Contextual Applications of Differentiation, about 10 minutes.
By the end of this lesson, you will be able to find the linearization of a function at a point, use it to approximate function values, and interpret the differential dy.
Local Linearity
If $f$ is differentiable at $x = a$, it looks linear when you zoom in close enough to $x = a$. The tangent line at $x = a$ is the best linear approximation to $f$ near that point.
The Linearization
The linearization of $f$ at $x = a$ is the tangent line equation:
$$L(x) = f(a) + f'(a)(x - a)$$
For $x$ close to $a$: $f(x) \approx L(x)$
This is the most important approximation tool in calculus — it's used everywhere in physics, engineering, and numerical computation.
Example 1: Approximating $\sqrt{4.1}$
We want to approximate $\sqrt{4.1}$, which is close to $\sqrt{4} = 2$.
Let $f(x) = \sqrt{x}$, and use $a = 4$ (a nearby "nice" value).
$f(4) = 2$, $f'(x) = \frac{1}{2\sqrt{x}}$, $f'(4) = \frac{1}{4}$
$L(x) = 2 + \frac{1}{4}(x - 4)$
$\sqrt{4.1} \approx L(4.1) = 2 + \frac{1}{4}(0.1) = 2 + 0.025 = 2.025$
Actual value: $\sqrt{4.1} \approx 2.0248$ — very close!
Example 2: Approximating $e^{0.1}$
$f(x) = e^x$, $a = 0$ (since $e^0 = 1$).
$f(0) = 1$, $f'(0) = 1$
$L(x) = 1 + x$
$e^{0.1} \approx 1 + 0.1 = 1.1$
Actual value: $e^{0.1} \approx 1.10517$ — linear approximation is good for small values near 0.
The approximation $e^x \approx 1 + x$ for small x is used throughout physics and engineering.
Overestimate or Underestimate?
The linearization overestimates or underestimates based on concavity:
- If $f''(a) > 0$ (concave up near $a$): $L(x)$ underestimates $f(x)$
- If $f''(a) < 0$ (concave down near $a$): $L(x)$ overestimates $f(x)$
Why? The tangent line is below a concave-up curve and above a concave-down curve.
Differentials
For $y = f(x)$, the differential $dy$ approximates the change in y for a small change $dx$ in x:
$$dy = f'(x), dx$$
Interpretation: $dy$ is the change in y along the tangent line; $\Delta y$ is the actual change in y.