4.2 Linear Approximations and Differentials

Big idea

A derivative was built as the slope of a tangent line. This section turns that construction around and uses the tangent line as a replacement for the function. Near the point of tangency the two graphs are close, so any question about $f$ that only concerns a small neighborhood can be answered with a line instead, and lines are easy.

The payoff is numerical. You know $\sqrt{4}$ exactly and you do not know $\sqrt{4.1}$, but you do know the slope of the square root function at $4$. Walking $0.1$ along the tangent line from the known point lands very near the unknown value. The same move estimates $\sin(0.1)$, $(1.02)^{10}$, or any value close to a point where you can evaluate the function and its derivative by hand.

There is a second payoff in measurement. Physical inputs come with uncertainty, and you often want to know how far the uncertainty propagates into a computed output. Since the tangent line converts a small input change into an output change by multiplying by $f'$, the derivative is exactly the amplification factor. That is the language of differentials.

Both uses rest on the same fact: differentiability at a point means the graph flattens out to a line under magnification. The approximation is not merely close, it is close in a strong sense, with the error shrinking faster than the distance you moved.

Decoder

Near a point where $f$ is differentiable, the tangent line at that point is the best linear stand-in for $f$, and the gap between them shrinks faster than the distance from the point.

The word doing the work is “near.” Nothing here claims global accuracy. Move far from the base point and the line and the curve part company, and how fast they part is governed by the bending of $f$, which the line knows nothing about. Choosing a base point close to your target, and one where you can compute exactly, is most of the skill.

“Faster than the distance” is what makes the approximation worth having. If you halve the distance from the base point, the error falls by much more than half; for a smooth function it falls by roughly a factor of four, because the leading error term is proportional to the square of the step. That is why a tangent line estimate of $\sqrt{4.1}$ is good to four digits while the same line at $\sqrt{9}$ would be useless.

The classic mistake is forgetting that the tangent line sits on one definite side of the curve. Where $f$ is concave down the tangent line lies above the graph, so the estimate is too big; where $f$ is concave up it lies below and the estimate is too small. Knowing the sign of the error for free is worth the one extra thought.

Definitions and results

The linearization. If $f$ is differentiable at $a$, its linearization at $a$ is

$$ L(x) = f(a) + f'(a)(x - a) $$

and the linear approximation is $f(x) \approx L(x)$ for $x$ near $a$. The graph of $L$ is the tangent line at $\big(a, f(a)\big)$.

Tangent line. The tangent line to $y = f(x)$ at $x = a$ has slope $f'(a)$ and passes through $\big(a, f(a)\big)$. Its equation is the linearization written as a line.

Normal line. The normal line at the same point is perpendicular to the tangent, so when $f'(a) \neq 0$ its slope is $-1/f'(a)$ and its equation is

$$ y = f(a) - \frac{1}{f'(a)}(x - a) $$

If $f'(a) = 0$ the tangent is horizontal and the normal is the vertical line $x = a$.

Differentials. Write $dx$ for an independent change in the input and define the differential of the output by

$$ dy = f'(x)\,dx $$

Here $dy$ is the rise along the tangent line, while the true change $\Delta y = f(x + dx) - f(x)$ is the rise along the curve. The approximation $\Delta y \approx dy$ is the linear approximation in different clothing.

Propagated error. If a measured quantity $x$ carries an uncertainty $dx$, the computed quantity $f(x)$ carries an uncertainty of roughly $|f'(x)\,dx|$. This is an absolute error, in the units of $f$.

Relative and percentage error. Dividing by the value gives a unitless comparison:

$$ \frac{dy}{y} = \frac{f'(x)\,dx}{f(x)} $$

For a power $y = x^{n}$ this collapses to $dy/y = n\,(dx/x)$, so a $1$ percent error in a length becomes a $3$ percent error in a volume.

Sign of the error. Where $f$ is concave up near $a$, the tangent lies below the curve and the estimate undershoots. Where $f$ is concave down, the tangent lies above and the estimate overshoots.

Worked examples

Estimating a square root

Approximate $\sqrt{4.1}$.

Take $f(x) = \sqrt{x}$ and base point $a = 4$, because $f(4) = 2$ is exact and $4.1$ is close. Since $f'(x) = \frac{1}{2\sqrt{x}}$, the slope is $f'(4) = \frac14$, and

$$ L(x) = 2 + \frac{1}{4}(x - 4) $$

So $\sqrt{4.1} \approx 2 + \frac{1}{4}(0.1) = 2.025$.

Check by squaring: $2.025^2 = 4.100625$, slightly more than $4.1$, so the estimate is slightly too large. That agrees with the concavity rule, since $\sqrt{x}$ is concave down and its tangent line sits above the graph. The true value is $2.02484\ldots$, so the error is about $1.6 \times 10^{-4}$.

A tangent line and a normal line

Find the tangent and normal lines to $f(x) = x^3 - 2x$ at $x = 1$.

Evaluate: $f(1) = 1 - 2 = -1$, so the point is $(1, -1)$. Differentiate: $f'(x) = 3x^2 - 2$, so $f'(1) = 1$.

The tangent has slope $1$:

$$ y = -1 + 1(x - 1) = x - 2 $$

The normal has slope $-1$:

$$ y = -1 - 1(x - 1) = -x $$

Check that the normal passes through the right point: at $x = 1$ it gives $y = -1$, matching. The two lines meet at $(1,-1)$ at a right angle, since the product of the slopes is $-1$.

Error in a measured volume

A cube is measured to have edge $10$ cm, with the measurement good to within $0.05$ cm. Estimate the resulting uncertainty in the computed volume.

The volume is $V = s^3$, so $dV = 3s^2\,ds$. With $s = 10$ and $ds = 0.05$,

$$ dV = 3(100)(0.05) = 15 $$

So the volume $1000$ cm$^3$ carries an uncertainty of about $15$ cm$^3$. In relative terms,

$$ \frac{dV}{V} = 3\cdot\frac{ds}{s} = 3(0.005) = 0.015 $$

A half percent uncertainty in the edge becomes a $1.5$ percent uncertainty in the volume. Cubing triples the relative error, which is the general rule for powers.

A trigonometric estimate

Approximate $\sin(0.1)$, with the angle in radians.

Take $f(x) = \sin x$ at $a = 0$: $f(0) = 0$ and $f'(0) = \cos 0 = 1$, so $L(x) = x$ and $\sin(0.1) \approx 0.1$. This is the small angle approximation, and it is nothing more than a tangent line at the origin.

Sine is concave down just to the right of $0$, so the estimate should be too large. It is: the true value is $0.099833\ldots$. The same reasoning at $a = 0$ gives $\cos x \approx 1$, because the cosine has a horizontal tangent there; that flat tangent is a hint that a good approximation of the cosine needs a curved correction, which later tools supply.

Practice

Start with the estimates. Pick a base point where the function and its derivative are exact, build the linearization, and evaluate it at the target.

Then the underlying line itself. These ask for the tangent line, the normal line, or the point where one of them meets a given target.

Quiz

Five items on linearizations, differentials and propagated error, and tangent and normal lines.