Which limit gives f'(x) for f(x) = x^2 + x?

\lim_{h\to 0}\frac{\bigl((x+h)^2+(x+h)\bigr)-(x^2+x)}{h}

\lim_{h\to 0}\frac{\bigl((x+h)^2+x\bigr)-(x^2+x)}{h}

\lim_{h\to 0}\frac{(x^2+x+h)-(x^2+x)}{h}

\lim_{h\to 0}\frac{(x+h)^2+(x+h)}{h}

Build a derivative limit

Builds on comparing average and instantaneous rates. 7 minutes.

Bacteria that double every hour

A culture holds \(2^t\) thousand bacteria after \(t\) hours. The orange point \(P\) shows the culture at 3 hours, \(2^3 = 8\) thousand. The point \(Q\) shows the culture a step \(h\) later, at \(3 + h\) hours.

For \(h = 1\), the culture grows from 8 to 16 thousand, so the green vertical leg shows a rise of 8. The line through \(P\) and \(Q\) is a secant line, and its slope is the rise divided by the step, \(8 \div 1 = 8\) thousand per hour.

A negative \(h\) puts \(Q\) to the left of \(P\). For \(h = -1\), the culture held 4 thousand an hour earlier, so the rise is \(4 - 8 = -4\) and the slope is \(-4 \div (-1) = 4\). The leg is red because \(Q\) is lower than \(P\).

Each \(h\) gives one value of the fraction \(\dfrac{2^{3+h} - 8}{h}\), called a difference quotient. The rate at 3 hours is the number that these values approach as \(h\) approaches 0.

Any function, any point

The function can be any \(f\), and \(P\) can be at any input \(a\). The point \(Q\) is at the input \(a + h\), so the step from \(P\) to \(Q\) is \(h\).

The rise is the output at \(Q\) minus the output at \(P\), \(f(a+h) - f(a)\). Subtract in the same order as the inputs, because the rise and the step both run from \(P\) to \(Q\).

As \(h\) shrinks, \(Q\) moves toward \(P\), and the secant line turns toward the tangent line. The tangent line touches the curve at \(P\) and points in the direction of the curve there.

The derivative \(f'(a)\) is the number that the secant slopes approach as \(h\) approaches 0.

Notation

At a fixed input \(a\), the difference quotient has three pieces: the output \(f(a+h)\) at \(Q\), the output \(f(a)\) at \(P\), and the step \(h\).
\[\frac{f(a+h)-f(a)}{h}\]
The derivative at \(a\) is the limit of this quotient as \(h\) approaches 0.
\[\cy{f'(a)}=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\]
The step \(h\) approaches 0 but is never 0, because the quotient divides by \(h\).
At a variable input \(x\), replace \(a\) by \(x\). The limit is now a function of \(x\).
\[\cy{f'(x)}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\]
While \(h\) approaches 0, the input \(x\) stays fixed.
The same limit can use a second input \(x\) that approaches \(a\), with \(x - a\) in place of \(h\).
\[f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\]
The two forms agree because \(x = a + h\), so \(x - a\) is the step \(h\), and \(x\) approaches \(a\) as \(h\) approaches 0.

When it fails

One value of \(h\) gives only an estimate of the rate at 3 hours. With \(h = 0.1\) the quotient is about 5.74, and with \(h = -0.1\) it is about 5.36.

As \(h\) approaches 0 from either side, the quotients approach one number, the derivative. It is about 5.55, the slope of the tangent at \(P\).

Example 1: a limit at a point

Write the limit that gives \(f'(2)\) for \(f(x) = \ln x\). Do not find its value.

At \(P\), the input is \(a = 2\) and the output is \(f(2) = \ln 2\), the value that the quotient subtracts.
\[f(2)=\ln 2\]
Replace \(x\) by \(2 + h\) in the formula to find the output at \(Q\).
\[f(2+h)=\ln(2+h)\]
Put the pieces in the quotient and take the limit.
\[\cy{f'(2)=\lim_{h\to 0}\frac{\ln(2+h)-\ln 2}{h}}\]
Do not substitute \(h = 0\) here, because the quotient would be \(0 \div 0\).

Example 2: a limit at a variable input

Write the limit that gives \(f'(x)\) for \(f(x) = 3x^2 + 1\). Do not find its value.

Replace \(x\) by \(x + h\) everywhere in the formula, and put the new input in parentheses.
\[f(x+h)=3(x+h)^2+1\]
A tempting wrong move: Write the output at \(Q\) as the old output plus \(h\).
\[\cr{\frac{(3x^2+1+h)-(3x^2+1)}{h}}\]
In the figure, \(x = 1\) and \(h = 1\), so this rule gives the height \(4 + 1 = 5\) at \(x = 2\), but the curve is 13 high there. The red line to this wrong point has slope 1, while the secant through \(P\) and \(Q\) has slope 9.
Use \(f(x+h)\) from the first step, subtract \(f(x)\), divide by \(h\) and take the limit.
\[\begin{aligned}\cy{f'(x)}&=\cy{\lim_{h\to 0}}\\ &\cy{\frac{3(x+h)^2+1-(3x^2+1)}{h}}\end{aligned}\]

Practice

Write the limit that gives \(g'(x)\) for \(g(x) = \sin 2x\). Do not find its value.

The point \(Q\) is at the input \(x + h\), so replace every \(x\) in the formula by \(x + h\). The input of the sine becomes \(2(x+h)\), not \(2x + h\).

Solution

Replace \(x\) by \(x + h\) in the formula to find the output at \(Q\).
\[g(x+h)=\sin\bigl(2(x+h)\bigr)\]
Subtract \(g(x)\), divide by \(h\) and take the limit.
\[\cy{\begin{aligned} &g'(x) = \\ &\lim_{h\to 0}\frac{\sin\bigl(2(x+h)\bigr)-\sin 2x}{h} \end{aligned}}\]

Questions

Questions 1 and 2 use the graph of \(f(x) = (x-1)^3 + 2\), with \(P\) at \(x = 1\). Question 3 needs a calculation.

\(Q\) is the point at \(x = 3\), so \(h = 2\). What is the value of \(\dfrac{f(1+h)-f(1)}{h}\)?
- 4
- 8
- 5
Answer
4
\(Q\) is now the point at \(x = 0\), to the left of \(P\), so \(h = -1\). What is the value of \(\dfrac{f(1+h)-f(1)}{h}\)?
- 1
- −1
- 2
Answer
1
Which limit gives \(f'(x)\) for \(f(x) = x^2 + x\)?
- \(\displaystyle\lim_{h\to 0}\frac{\bigl((x+h)^2+(x+h)\bigr)-(x^2+x)}{h}\)
- \(\displaystyle\lim_{h\to 0}\frac{\bigl((x+h)^2+x\bigr)-(x^2+x)}{h}\)
- \(\displaystyle\lim_{h\to 0}\frac{(x^2+x+h)-(x^2+x)}{h}\)
- \(\displaystyle\lim_{h\to 0}\frac{(x+h)^2+(x+h)}{h}\)
Answer
\(\displaystyle\lim_{h\to 0}\frac{\bigl((x+h)^2+(x+h)\bigr)-(x^2+x)}{h}\)

Summary

You find \(f(a+h)\) by replacing \(x\) with \(a + h\), and you subtract \(f(a)\).

You divide by \(h\), because the quotient is the slope of the secant line from the input \(a\) to the input \(a + h\).

You take the limit as \(h\) approaches 0 to define the derivative at a point, or at a variable input \(x\).

This is one of four skills in topic 2.2 of Which & Why: AP Calculus.