Near the point of tangency a curve and its tangent line are almost the same. That makes the tangent line a cheap way to estimate values that are hard to compute. With the The Chain Rule we can do this for almost any function.
Key idea
The linearization of $f$ at $a$ is $L(x) = f(a) + f'(a)(x - a)$, and $f(x) \approx L(x)$ for $x$ near $a$. In differential form: if $x$ changes by $dx$, then $f$ changes by about $dy = f'(x)\,dx$.
Why it works
The derivative is the limit of $\frac{f(x) - f(a)}{x - a}$, so for $x$ near $a$ the quotient is close to $f'(a)$: $f(x) - f(a) \approx f'(a)(x - a)$. Rearranged, that is $f(x) \approx f(a) + f'(a)(x - a)$. The error shrinks faster than $x - a$ itself, which is what "the curve looks straight up close" means. The farther $x$ is from $a$, the worse the estimate.
Worked example: A square root
Estimate $\sqrt{26}$ with a linearization.
- Pick a nearby easy point: $a = 25$, $f(x) = \sqrt{x}$, $f(25) = 5$.
- $f'(x) = \frac{1}{2\sqrt{x}}$, so $f'(25) = \frac{1}{10}$.
- $L(26) = 5 + \frac{1}{10}(26 - 25) = 5.1$.
- The true value is $5.0990\ldots$: off by about $0.001$.
Answer: $\sqrt{26} \approx 5.1$
Worked example: A cube root
Estimate $\sqrt[3]{8.06}$.
- $a = 8$, $f(x) = x^{1/3}$, $f(8) = 2$, $f'(x) = \frac{1}{3}x^{-2/3}$, so $f'(8) = \frac{1}{3 \cdot 4} = \frac{1}{12}$.
- $L(8.06) = 2 + \frac{1}{12}(0.06) = 2.005$.
Answer: $\sqrt[3]{8.06} \approx 2.005$ (true value about $2.004988$)
Worked example: Measurement error
A disk's radius is measured as 10 cm with a possible error of 0.1 cm. Estimate the possible error in its area.
- $A = \pi r^2$, so $dA = 2\pi r\,dr$.
- $dA = 2\pi(10)(0.1) = 2\pi \approx 6.3$ cm².
Answer: About $2\pi \approx 6.3$ cm²
Common mistake
Using a base point that is far away, or one where the values are not easy. The method only helps if $f(a)$ and $f'(a)$ are known exactly and $x$ is close to $a$. Check: also note which side the error lies on. For $\sqrt{x}$, which bends downward, the tangent line lies above the curve, so $L$ overestimates.
Physicists use these approximations constantly: $\sin x \approx x$ and $(1 + x)^n \approx 1 + nx$ for small $x$ are both linearizations at 0.