Can I use derivatives to estimate a value like √26 quickly without a calculator?
Yes — linear approximation.
Yes — linear approximation. Take f(x) = √x near a known point x = 25, where f(25) = 5 and f'(x) = 1/(2√x) = 1/10. Then f(26) ≈ f(25) + f'(25)×(1) = 5 + 0.1 = 5.1, which is very close to the true 5.099. The derivative gives the slope of the tangent line, and for a small step δx the tangent is a great stand-in for the curve, so f(x+δx) ≈ f(x) + f'(x)δx.
Key point
f(x+δx) ≈ f(x) + f'(x)δx — the tangent line approximates the curve for small δx.
Common mistake
using a far-away base point where the tangent no longer matches the curve.
Memory tip
pick a nearby 'nice' point; the tangent estimates the rest.
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