Linearization and Differentials

Zoom in far enough on any differentiable curve and it looks like a straight line — its tangent. The linearization \(L(x) = f(a) + f'(a)(x-a)\) cashes that in: replace a hard function with an easy line and get a fast, accurate estimate near the base point.

By the end you'll be able to build \(L(x)\) at a well-chosen base point, use it to estimate values like \(\sqrt{9.1}\) by hand, and use the differential \(dy = f'(x)\,dx\) to estimate a change — knowing which way the error leans.

Predict: \(f(x)=\sqrt{x}\) is concave down — it bends under its tangent lines. Will the tangent-line estimate land above or below the true value? Drag x away from the base point a and check — and watch what the growing gap does to the error.

The dashed line is the tangent at \(x=a\): its height L(x) is the estimate, the curve's height f(x) is the truth, and the shaded band between them is the approximation error. Move a to re-anchor the tangent; drag either dot (or use the x slider) to move the evaluation point. Hover or focus a dot for exact values.

L(x) = 3.500 at x = 12.00 · true f(x) = 3.464 · error = 0.036 (tangent overestimates)

f(x) = √x and its tangent line at x = a — the shaded band is the error |L(x) − f(x)|
f(x) = √x tangent L(x) error band

The linearization \(L(x) = f(a) + f'(a)(x-a)\) uses the tangent line to approximate \(f\) near \(a\), and the differential \(dy = f'(x)\,dx\) estimates the resulting change — with an error that shrinks faster than the step itself.

The tangent line as a stand-in

Formal

\(L(x) = f(a) + f'(a)(x-a)\) is the unique line through \((a, f(a))\) with slope \(f'(a)\) — start at the known height \(f(a)\), then walk the run \((x-a)\) at the tangent's rise-per-run. Differentiability guarantees \(f(x) - L(x) \to 0\) faster than \(x - a\) does, which is exactly why the shaded band in the diagram collapses so quickly as \(x\) slides toward \(a\). Farther out the guarantee expires: the error grows with \(|x-a|\), and faster where the curve bends sharply (large \(|f''|\)). Which side the error lands on is set by concavity — concave down means the curve ducks under the tangent, so L(x) overestimates; concave up flips it to an underestimate.

Applied

This is how you estimate \(\sqrt{9.1}\) without a calculator: anchor at the friendly neighbor \(a = 9\), where \(f(9)=3\) and the slope is easy, and ride the tangent the last 0.1. The same move powers error propagation in lab work — if a measured length carries a small uncertainty \(dx\), the differential \(dy = f'(x)\,dx\) tells you the resulting uncertainty in any computed quantity — and it's the first step of Newton's method for solving equations. The craft is choosing \(a\): close to the target and easy to evaluate.

Differentials: naming the predicted change

Write \(dy\) \(= f'(x)\,dx\) for the change the tangent line predicts when \(x\) moves by \(dx\). The true change is \(\Delta y\) \(= f(x+dx) - f(x)\). For small \(dx\) they nearly agree, and their gap — the approximation error — shrinks faster than \(dx\) itself: halve the step and the error drops by roughly a factor of four. So the recipe is: pick a base point \(a\) where \(f(a)\) and \(f'(a)\) are easy, form \(L(x)\), and either evaluate \(L\) at the target or add \(dy = f'(a)\,dx\) to \(f(a)\) — same arithmetic, two vocabularies.

Worked example

Estimate \(\sqrt{9.1}\). Take \(f(x)=\sqrt{x}\) and base point \(a=9\), so \(f(9)=3\). Writing \(\sqrt{x}=x^{1/2}\), the power rule gives \(f'(x)=\tfrac{1}{2}x^{-1/2}=1/(2\sqrt{x})\), so \(f'(9)=1/6\). Then \(L(x)=3+\tfrac{1}{6}(x-9)\), and \(L(9.1)=3+\tfrac{0.1}{6}\approx \) 3.0167. The true value is 3.0166 — off by less than 0.0001, and (concave down!) the estimate sits just above the truth.

Your turn

Estimate \(\sqrt{4.05}\). Take \(a=4\), so \(f(4)=2\) and \(f'(4)=1/(2\sqrt{4})=1/4\). Then \(L(x)=2+\tfrac{1}{4}(x-4)\), so \(L(4.05)=2+\tfrac{1}{4}(0.05)= \) ____

Reveal the answer

\(L(4.05)=2+0.0125= \) 2.0125 — the true value is 2.01246, so the tangent overshoots by about 0.00004. Set a = 4 and x = 4.05 in the explorer above: the error band all but vanishes at this tiny step.

More info — why zooming in makes every smooth curve look straight

Differentiability at \(a\) literally means the difference quotient \(\frac{f(x)-f(a)}{x-a}\) settles down to a single number \(f'(a)\) as \(x \to a\). Rearranged, that says \(f(x) \approx f(a) + f'(a)(x-a)\) with an error that vanishes faster than \(x-a\) — the linearization is nothing more than the derivative's definition read as an approximation instead of a limit. It's also the first-order Taylor approximation: keep the constant and linear terms, drop the bends. The OpenStax chapter in Dive deeper below carries this straight into the differentials notation used in the sciences.

Check your understanding

Question 1 of 4

Use the linearization of \(f(x)=\sqrt{x}\) at \(a=25\) to estimate \(\sqrt{24.8}\).

Question 2 of 4

\(f\) is concave down near \(a\) (like \(\sqrt{x}\)). What can you say about the tangent-line estimate \(L(x)\) for \(x\) near \(a\)?

Question 3 of 4

To linearize \(f(x)=\sqrt[3]{x}\) at \(a=8\) you first need \(f'(8)\). Write \(\sqrt[3]{x}=x^{1/3}\) and apply the power rule: \(f'(8)=\) ?

Question 4 of 4

Let \(y=x^2\) and let \(x\) change from 3 to 3.1. The differential gives \(dy=f'(3)\,dx=6(0.1)=0.6\), while the true change is \(\Delta y=3.1^2-3^2=0.61\). Which statement is right?

Recap

  • The linearization of \(f\) at \(a\) is \(L(x) = f(a) + f'(a)(x-a)\) — the tangent line used as a stand-in for \(f\), with \(L(x)\approx f(x)\) for \(x\) near \(a\).
  • The differential \(dy = f'(x)\,dx\) is the tangent line's predicted change; the true change is \(\Delta y = f(x+dx)-f(x)\), and their gap shrinks faster than \(dx\).
  • Pick a base point that is close to the target AND easy to evaluate; accuracy degrades as \(|x-a|\) grows, faster where \(|f''|\) is large.
  • Concavity sets the bias: concave down → the tangent overestimates; concave up → it underestimates.

Dive deeper

Sources

  • Linearization and Differentials