Linear approximation is the practical payoff of the derivative: at a point where $f$ is differentiable, the graph of $f$ is indistinguishable from its tangent line, so a hard function can be replaced locally by an easy linear one. This topic turns that idea into a formula (the linearisation), a notation for small changes (differentials), and — the part students skip — a way to say how wrong the answer is and in which direction. It is the first-order Taylor polynomial you will meet again in integral calculus, and it is the engine behind every "propagation of measurement error" calculation in physics and engineering.

The linearisation and the tangent-line approximation

Definition. If $f$ is differentiable at $a$ , the linearisation of $f$ at $a$ is $L(x) = f(a) + f'(a)(x-a).$ The graph $y = L(x)$ is exactly the tangent line to $y=f(x)$ at the point $\left(a, f(a)\right)$ . The linear approximation (tangent-line approximation) is the statement $f(x) \approx L(x) \qquad \text{for } x \text{ near } a.$

Two hypotheses carry all the weight: $f'(a)$ must exist, and $x$ must be near $a$ . In fact differentiability at $a$ is precisely the statement that $f(x) = f(a) + f'(a)(x-a) + E(x)(x-a), \qquad E(x) \to 0 \text{ as } x \to a,$ so the error is not merely small — it is small compared with $x-a$ . Nothing here promises anything far from $a$ : $\sin x \approx x$ is excellent at $x = 0.05$ and useless at $x = 3$ .

Choosing $a$ . Pick the nearest point at which $f(a)$ and $f'(a)$ are computable exactly: a perfect square or cube, $x=1$ , $x=0$ , $\pi/6$ , $\pi/4$ . To estimate $\sqrt{26}$ use $a=25$ , not $a=16$ .

Standard linearisations at $a = 0$

$f(x)$	$L(x)$ near $0$	Sign of $f''$ near $0$	$L$ over- or under-estimates
$(1+x)^{k}$ (needs $x>-1$ )	$1+kx$	sign of $k(k-1)$	over if $0<k<1$ ; under if $k<0$ or $k>1$ ; exact if $k=0$ or $k=1$
$\sqrt{1+x}$	$1+\dfrac{x}{2}$	negative	over
$\dfrac{1}{1+x}$	$1-x$	positive	under
$e^{x}$	$1+x$	positive	under
$\ln(1+x)$	$x$	negative	over
$\sin x$	$x$	opposite sign to $x$	over for $x>0$ , under for $x<0$
$\tan x$	$x$	same sign as $x$	under for $x>0$ , over for $x<0$
$\arctan x$	$x$	opposite sign to $x$	over for $x>0$ , under for $x<0$
$\cos x$	$1$	negative	over

Note the last row: $f'(0)=0$ , so the linearisation of $\cos$ at $0$ is the constant $1$ . That is not a failure — it is why $\cos x \approx 1$ is unusually accurate for small $x$ (the error is of order $x^{2}$ , not $x$ ).

Differentials

Definition. Let $y = f(x)$ with $f$ differentiable. The symbol $dx$ is an independent variable: it may be given any real value. The differential $dy$ is then defined by $dy = f'(x) dx.$

Compare it with the true change. Fix $x$ and a step $\Delta x$ , and set $dx = \Delta x$ . Then

$\Delta y = f(x + \Delta x) - f(x)$ is the actual rise along the curve;
$dy = f'(x)\Delta x$ is the rise along the tangent line.

So $\Delta y \approx dy$ , and $\Delta y - dy \to 0$ faster than $\Delta x$ does. Evaluating the differential at the base point — the derivative taken at $a$ , with $dx = x-a$ — recovers the linearisation: $f(x) \approx f(a) + dy = L(x)$ . Once $dy$ and $dx$ are separate objects, the Leibniz symbol $\dfrac{dy}{dx} = f'(x)$ really is a quotient — which is why chain-rule and separable-equation notation behaves so well.

Estimating a small change: the recipe

Identify the quantity $y = f(x)$ and the base point $a$ where $f(a)$ , $f'(a)$ are exact.
Write $dx = x - a$ (the small change or the measurement error).
Compute $dy = f'(a) dx$ .
Report $f(x) \approx f(a) + dy$ , or "the change in $y$ is about $dy$ ".
Say whether it is an over- or under-estimate (concavity) and, if asked, bound the error.

Absolute, relative and percentage error

For a quantity $q$ with estimated error $dq$ :

absolute error $= \left| dq \right|$ (carries units);
relative error $= \dfrac{\left| dq \right|}{\left| q \right|}$ (dimensionless);
percentage error $= 100 \cdot$ relative error.

Power rule for error propagation. If $y = Cx^{n}$ with $C \ne 0$ constant, and $x \ne 0$ (so neither denominator below vanishes), then $dy = Cnx^{n-1} dx$ , so $\frac{dy}{y} = \frac{Cnx^{n-1} dx}{Cx^{n}} = n \cdot \frac{dx}{x}.$ The percentage error in $y$ is about $\left| n \right|$ times the percentage error in $x$ . Hence a $1\%$ error in a radius gives about $2\%$ in an area and $3\%$ in a volume, while $T = 2\pi\sqrt{L/g}$ turns a $1\%$ error in $L$ into only $0.5\%$ in $T$ .

Over- or under-estimate: read it off the concavity

Theorem. Suppose $f''$ exists on an open interval $I$ containing $a$ .

If $f''(x) \ge 0$ for all $x$ in $I$ ( $f$ concave up), then $f(x) \ge L(x)$ on $I$ : the tangent line lies below the graph, so $L$ under-estimates.
If $f''(x) \le 0$ for all $x$ in $I$ ( $f$ concave down), then $f(x) \le L(x)$ on $I$ : the tangent lies above, so $L$ over-estimates.

It makes no difference whether $x > a$ or $x < a$ — only the sign of $f''$ matters. If $f''$ changes sign between $a$ and $x$ (an inflection point in between), no conclusion is available.

Error bounds

Theorem (Taylor's inequality, first order). Suppose $f'$ is continuous on the closed interval with endpoints $a$ and $x$ , and $f''$ exists on the open interval between them. Then there is a number $c$ strictly between $a$ and $x$ with $f(x) - L(x) = \frac{f''(c)}{2}(x-a)^{2}.$ Consequently, if $\left| f''(t) \right| \le M$ for every $t$ between $a$ and $x$ , then $\left| f(x) - L(x) \right| \le \frac{M}{2}(x-a)^{2}.$

Three consequences worth memorising: (i) $M$ must bound $\left| f'' \right|$ across the whole interval, not just at $a$ ; (ii) any valid $M$ gives a valid bound, so a clean over-estimate of $M$ is perfectly acceptable; (iii) the bound is quadratic in the step, so halving $\left| x-a \right|$ quarters it.

Worked example 1 — a cube root, with direction and error bound

Estimate $\sqrt[3]{8.06}$ , decide whether the estimate is too big or too small, and bound the error.

Take $f(x) = x^{1/3}$ and $a = 8$ (the nearest perfect cube).

\begin{aligned} f(8) &= 2, \\ f'(x) &= \frac{1}{3}x^{-2/3}, \qquad f'(8) = \frac{1}{3 \cdot 4} = \frac{1}{12}, \\ L(x) &= 2 + \frac{1}{12}(x-8). \end{aligned}

With $x = 8.06$ , so $dx = 0.06$ : $L(8.06) = 2 + \frac{0.06}{12} = 2 + 0.005 = 2.005.$

Direction. $f''(x) = -\dfrac{2}{9}x^{-5/3} < 0$ for $x>0$ , so $f$ is concave down and $2.005$ is an over-estimate.

Bound. For $t$ in $[8, 8.06]$ we have $t \ge 8$ , hence $t^{5/3} \ge 8^{5/3} = 32$ and $\left| f''(t) \right| = \frac{2}{9}t^{-5/3} \le \frac{2}{9} \cdot \frac{1}{32} = \frac{1}{144} = M.$ Therefore $\left| \sqrt[3]{8.06} - 2.005 \right| \le \frac{1}{288}(0.06)^{2} = \frac{0.0036}{288} = 1.25 \times 10^{-5}.$ Combining with the direction: $2.0049875 \le \sqrt[3]{8.06} \le 2.005$ . (The true value is $2.0049876$ , so both statements are correct.)

Worked example 2 — measurement error

The radius of a circular disk is measured as $24$ cm with a maximum error of $0.02$ cm. Estimate the resulting maximum error, relative error and percentage error in the computed area.

$A = \pi r^{2}$ , so $dA = 2\pi r dr$ . With $r = 24$ and $\left| dr \right| \le 0.02$ : $\left| dA \right| \le 2\pi (24)(0.02) = 0.96\pi \approx 3.02 \text{ cm}^{2}.$ Relative error: $\frac{\left| dA \right|}{A} \le \frac{0.96\pi}{\pi (24)^{2}} = \frac{0.96}{576} = \frac{1}{600} \approx 0.00167,$ i.e. about $0.167\%$ . The shortcut $\dfrac{dA}{A} = 2\dfrac{dr}{r} = \dfrac{2(0.02)}{24} = \dfrac{1}{600}$ gives the same thing with no arithmetic. (Exactly: $A(24.02) - A(24) = 0.9604\pi \approx 3.017$ , against the estimate $3.016$ .)

Common mistakes

Wrong	Right
$L(x) = f(a) + f'(x)(x-a)$	$L(x) = f(a) + f'(a)(x-a)$ — the derivative is frozen at $a$ , so $L$ is genuinely linear in $x$
Estimating $\sin 31^{\circ}$ with $dx = 1$	Convert to radians first: $dx = \pi/180 \approx 0.01745$
Using $a = 8.06$ to estimate $\sqrt[3]{8.06}$	Then $x-a=0$ and you learn nothing; $a$ must be a nearby point you can evaluate exactly
" $x > a$ , so the estimate is too small"	The side of $a$ is irrelevant; the sign of $f''$ decides over- vs under-estimate
Treating $\Delta y$ and $dy$ as equal	$dy$ is the tangent-line rise; $\Delta y - dy$ is nonzero. Differentiability alone gives only $\Delta y - dy \to 0$ faster than $\Delta x$ ; the familiar order $(\Delta x)^{2}$ needs $f''$ to exist and be bounded near $x$ (for $f(x)=x^{4/3}$ at $x=0$ the gap is $(\Delta x)^{4/3}$ , far bigger than $(\Delta x)^{2}$ )
Writing $dy = f'(x)$	$dy = f'(x) dx$ — without the $dx$ the equation is dimensionally meaningless
Taking $M = \left	f''(a) \right
Reporting $0.00167$ as "the percentage error"	Percentage error is $100 \times$ the relative error: $0.167\%$
Linearising $\left	x \right

Navigation

Linear Approximation and Differentials

The linearisation and the tangent-line approximation

Standard linearisations at $a = 0$

Differentials

Estimating a small change: the recipe

Absolute, relative and percentage error

Over- or under-estimate: read it off the concavity

Error bounds

Worked example 1 — a cube root, with direction and error bound

Worked example 2 — measurement error

Common mistakes

Key terms

Practice Problems

Navigation

The linearisation and the tangent-line approximation

Standard linearisations at a=0a = 0a=0

Differentials

Estimating a small change: the recipe

Absolute, relative and percentage error

Over- or under-estimate: read it off the concavity

Error bounds

Worked example 1 — a cube root, with direction and error bound

Worked example 2 — measurement error

Common mistakes

Key terms

Practice Problems

Standard linearisations at $a = 0$