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

The Derivative of 1/t

1/t is not a polynomial, so its quotient cannot be expanded. It can still be simplified exactly, by putting 1/(a + h) − 1/a over a common denominator. The h cancels, leaving −1/(a(a + h)), and as h shrinks the denominator heads for a², so the derivative is −1/a².

Open the interactive lesson Practice

1/(a + h) − 1/a = −h/(a(a + h)) the quotient: −1/(a(a + h)) a = 2 gives −1/6, −1/5, −2/9, −4/17 → −1/4 f′(a) = −1/a² negative, because 1/t falls

The derivative of 1/t

Core ideas

What this lesson establishes

Three ideas: combining the fractions, cancelling, and reading the number off a formula and a column together.

Put the difference over a common denominator

1/(a + h) − 1/a is a subtraction of fractions, and a common denominator is a(a + h). The numerator becomes a − (a + h) = −h, so the difference is −h/(a(a + h)). The h has been exposed as a factor of the numerator, which is the same event as in the polynomial case.

Cancel the h, and the quotient is a fraction in h

Dividing by h leaves −1/(a(a + h)). This is not a polynomial in h, but it is exact, and it has a value at h = 0 because the zero has been cancelled out of the denominator. The column of quotients is this fraction evaluated at h = 1, 1/2, 1/4, ….

The number it approaches is −1/a²

As h shrinks, a + h heads for a, so the denominator a(a + h) heads for a² and the quotient for −1/a². The column in the lab and the formula agree: at a = 2 the column heads for −1/4. The sign is negative because 1/t falls as t rises.

The material

Two fractions, one subtraction, one cancelled h

The algebra once in general, the column that confirms it, and why the answer cannot be positive or a logarithm.

The function is f(t) = 1/t, so f(a + h) = 1/(a + h). The quotient has a fraction inside a fraction, and the way through is to subtract the two fractions first.

f(a + h) − f(a) = 1/(a + h) − 1/a = (a − (a + h)) / (a·(a + h)) = −h / (a·(a + h)) divide by h: −1 / (a·(a + h))

The h in the numerator is the whole point, as in the polynomial lesson: once it is divided out, the quotient has no zero in its denominator. At a = 2 and h = 1/4 the quotient is −1/(2·(9/4)) = −2/9, and the long way gives the same: 1/(9/4) − 1/2 = 4/9 − 1/2 = −1/18, and dividing by 1/4 gives −2/9.

f(t) = 1/t, a = 2 h quotient −1/(2·(2 + h)) 1 −1/6 1/2 −1/5 1/4 −2/9 1/8 −4/17 1/16 −8/33 1/32 −16/65

The column heads for −1/4. Each entry is −1/(2·(2 + h)), and the denominator 2·(2 + h) heads for 4. The gaps from −1/4 shrink as the step does, and, as before, no row reaches −1/4.

Theorem · The derivative of 1/t

For every place a ≠ 0, the quotients of f(t) = 1/t are −1/(a(a + h)), and they approach −1/a² as h → 0: f′(a) = −1/a².

The quotient was derived exactly above, by algebra. What is left is the claim that −1/(a(a + h)) approaches −1/a² as h shrinks.

The denominator a(a + h) = a² + a·h, which differs from a² by a·h, and that can be made as small as you like by taking h small. So the quotient is as near −1/a² as you choose, and the lab's column, which shows several steps, illustrates it. The approach is the same claim as for a polynomial, and here it is read off a formula instead of a constant term.

□

Two other values of a

At a = 1 the quotient is −1/(1 + h), so the column is −1/2, −2/3, −4/5, −8/9, …, and it heads for −1. At a = 1/2 the quotient is −1/((1/2)·(1/2 + h)), the column starts at −4/3 and heads for −4. Both agree with −1/a²: at a = 1/2, a² = 1/4 and −1/(1/4) = −4. The rate is steeper where t is smaller, which fits the graph of 1/t, falling fast near 0 and flattening out for large t.

The tangent line from the last lesson uses this. At a = 1 the slope is −1, the point is (1, 1), and the line is y = −t + 2.

Interactive

Difference quotients

Average rates as exact fractions

Method

Differentiating 1/t by hand

The same four moves as the polynomial case, with fractions in place of an expansion.

Write f(a + h) − f(a) as two fractions

1/(a + h) − 1/a. Do not divide by h yet.

Combine over a common denominator

The common denominator is a(a + h). The numerator is a − (a + h) = −h.

Divide by h and cancel

The h in the numerator cancels the h in the divisor, leaving −1/(a(a + h)).

Let the step shrink in the formula

The denominator a(a + h) heads for a², so the quotient heads for −1/a². Check with a column: the fractions should agree.

Fractions and the derivative of 1/t

Score 0 / 0

Choose the answer you can justify.

1/t at 2, one step by hand

The place is a = 2 and the step is h = 1/4, so a + h = 9/4. Both routes are shown.

the long way: 1/(9/4) = 4/9 4/9 − 1/2 = 8/18 − 9/18 = −1/18 (−1/18)/(1/4) = −4/18 = −2/9 the formula: −1/(a·(a + h)) = −1/(2·(9/4)) = −2/9 as h gets small, a·(a + h) → 4, quotient → −1/4

Both routes give −2/9. The lab's column for a = 2 has this value in its third row, and the rows after it head for −1/4, which is −1/a² at a = 2.

For a rehearsal, find the derivative of 1/t at a = 3 from the formula, and check that the quotient at h = 1 is −1/12.

Common mistakes

Taking the derivative of 1/t to be 1 or ln tAt a = 1 the quotients of 1/t are −1/2, −2/3, −4/5, …, every one negative, because 1/t falls from 1 to 1/2 as t goes from 1 to 2. No positive number can be where they head, so the derivative is not 1, and ln t rises, so its slope would be positive too. It is −1 at a = 1, and −1/a² in general.

Applying a polynomial rule to a fraction1/t is not t to a positive power, so the expansion of the polynomial lesson does not apply. The cancelling is of a different kind: the h appears when the two fractions are combined. Skipping the combination leaves a fraction over h with no visible factor to cancel.

Forgetting that a cannot be zero1/t has no value at 0, so there is no quotient and no derivative there, and −1/a² is undefined at a = 0. The lab refuses a place of 0, or a place where a + h lands on 0, instead of printing something.

Check your understanding

✓

Finish when you can derive the quotient of 1/t by combining fractions and state its derivative at a place.

You should be able to write 1/(a + h) − 1/a as one fraction, cancel the h, evaluate the result at a given step, and say what the column of such quotients approaches.

Note: Two kinds of function now have a derivative at every place: a polynomial and 1/t. The next lesson stops computing one place at a time and treats the derivative as a function of its own, in “The Derivative as a Function”.