Lesson · Rates of Change and the Derivative · The derivative at a point

The Derivative at a Point and the Tangent Line

The number a column of quotients approaches at a is the derivative f′(a). The line through (a, f(a)) with that slope is the tangent line, y = f(a) + f′(a)·(t − a). It is not a line that touches once and says nothing about the rest: it is the best straight-line estimate near a, and its error at a + h is a fraction the lab prints exactly.

Open the interactive lesson Practice

f′(a) = what the quotients approach tangent: y = f(a) + f′(a)·(t − a) estimate at a + h: f(a) + f′(a)·h error = f(a + h) − estimate, exactly t² at 2, h = 1/10: error 1/100 = h²

The derivative at a point and its tangent line

Core ideas

What this lesson establishes

Three ideas: the name for the number, the line it builds, and how far the line can be trusted.

The derivative at a is the number the quotients approach

Write f′(a) for it, read “f prime of a”. It is a single number for each place a, not a family. For t² at 2 the quotient is 4 + h, the column heads for 4, and f′(2) = 4. That is a claim, as before, and for a polynomial it is backed by the constant term of the simplified quotient.

The tangent line is the line with that slope through the point

It passes through (a, f(a)) with slope f′(a), so its equation is y = f(a) + f′(a)·(t − a). For t² at 2 that is y = 4 + 4·(t − 2) = 4t − 4. It is the limit of the secant lines as the step shrinks, which is why its slope is the derivative.

Near a, the line estimates the function, and the error is computable

At t = a + h the line gives f(a) + f′(a)·h, while the function gives f(a + h). Their difference is the error of the linear approximation. For a polynomial it is an exact fraction, and it shrinks faster than h does, which is what makes the line useful.

The material

A number, a line, and the error of using it

The definition, the tangent line for a parabola, the exact error, and why “touches once” is the wrong picture.

Definition · The derivative at a point

The derivative of f at a is the number the difference quotients (f(a + h) − f(a))/h approach as h → 0. It is written f′(a).

The tangent line to the graph of f at a is the line through (a, f(a)) with slope f′(a): y = f(a) + f′(a)·(t − a).

Take f(t) = t² and a = 2. The quotient is ((2 + h)² − 4)/h = 4 + h, so the derivative is f′(2) = 4, the constant term. The point on the graph is (2, 4), and the tangent line is y = 4 + 4·(t − 2), which simplifies to y = 4t − 4.

Using the line to estimate

Move a tenth to the right, to t = 21/10. The line gives 4·(21/10) − 4 = 22/5, which is 440/100. The function gives (21/10)² = 441/100. The estimate is short by 1/100.

f(t) = t², a = 2, h = 1/10 f(2) = 4 f′(2) = 4 line at 21/10: 4 + 4·(1/10) = 22/5 = 440/100 function at 21/10: 441/100 error: 1/100

The error is exactly 1/100, which is h². That is no accident for this function. The function at a + h is 4 + 4h + h² and the line is 4 + 4h, so the difference is the h² term that the line leaves out. Halve the step to 1/20 and the error is 1/400: a quarter of it, not a half. The estimate gets better faster than the step gets smaller.

Example · A cube at 1

For f(t) = t³ at a = 1 the quotient is 3 + 3h + h², so f′(1) = 3 and the tangent line is y = 1 + 3·(t − 1) = 3t − 2.

At h = 1/10 the line gives 13/10 and the function gives 1331/1000. The error is 31/1000, which is 3h² + h³: the terms of the expansion that the line leaves out.

Example · A function that is not a polynomial

For f(t) = 1/t at a = 1 the lab reports a slope of −1, which the next lesson derives. The tangent line is y = 1 − (t − 1) = −t + 2.

At h = 1/2 the line gives 1/2 and the function gives 2/3, so the error is 1/6. The line is too low here because 1/t bends upward away from its tangent.

Touching is the wrong picture

A tangent is often described as a line that touches the graph at one point. That is true of a circle and misleading here. The tangent line to t² at 2 meets the parabola at (2, 4) and nowhere else, but that is not what defines it. It is defined by the slope, and its usefulness is in the nearby values it estimates: 22/5 against 441/100, off by 1/100, at a point a tenth of a unit away.

The error is always printed exactly in the lab, and for a polynomial it is made of the terms of f(a + h) that contain h² or more. That is why the estimate improves faster than the step shrinks, and it is the whole reason a derivative is useful.

Interactive

The tangent line

The linear approximation and its exact error

Method

From a quotient to a tangent line

Five moves. The first three find the derivative, the last two use it.

Simplify the quotient at the place

Expand and cancel the h, as in the polynomial lesson, with a in the place of t.

Read off the constant term

It is f′(a), the number the quotients approach.

Write the line

y = f(a) + f′(a)·(t − a): the point (a, f(a)) and the slope f′(a).

Estimate at a + h

The line gives f(a) + f′(a)·h. Compute it as a single fraction.

Subtract to find the error

True value minus estimate: f(a + h) − (f(a) + f′(a)·h). A positive error means the line lies below the function there.

Slope, line and error

Score 0 / 0

Choose the answer you can justify.

t² at 2, a tenth to the right

The function is f(t) = t², the place is a = 2 and the step is h = 1/10. Each number is an exact fraction.

f(2) = 4 quotient: ((2 + h)² − 4)/h = 4 + h, so f′(2) = 4 tangent line: y = 4 + 4·(t − 2) = 4t − 4 line at 21/10: 4·(21/10) − 4 = 22/5 = 440/100 function at 21/10: (21/10)² = 441/100 error: 441/100 − 440/100 = 1/100 = h²

The line is a good estimate: it is within one part in 441 of the true value. The error is the h² term that the line leaves out.

For a rehearsal, find the tangent to t² at 3 and use it at t = 31/10. The slope is 6, the estimate is 9 + 6/10 = 48/5, and the error should again be h².

Common mistakes

Thinking the tangent “touches at one point only”, so it says nothing about nearby valuesThe tangent to t² at 2 meets the parabola only at (2, 4), and yet at t = 21/10 it estimates 22/5 against the true 441/100, an error of 1/100. Touching once is not the definition; the slope is, and the line is valuable because the error falls as h² while the step falls as h.

Using the tangent far from its pointThe error is exact and it grows: at h = 1 the error for t² at 2 is 1, and at h = 10 it is 100. The line is an estimate near a, and the error formula says how near.

Confusing the derivative with a value of the functionf′(2) is the slope, 4, and f(2) is the height, also 4 here only by coincidence. For t³ at 1 the slope is 3 and the height is 1. The tangent line uses both: the height as the starting point and the slope as the rate.

Check your understanding

✓

Finish when you can find f′(a) for a polynomial, write the tangent line, and compute the exact error of the estimate at a + h.

You should be able to simplify the quotient at a place, read the derivative off as the constant term, write the line, evaluate it at a + h, and subtract from the true value to get a fraction.

Note: Every derivative so far came from a quotient that simplifies to a polynomial in h. One function does not do that, and it is worth doing by hand: “The Derivative of 1/t”.