The definition of the derivative, question 24

Topic: The definition of the derivative
Difficulty: Challenging
Suggested time: 6 min
Marks: 3
Theory: UBC Math 100 §3.3 CLP-1 §2.2Key results

Two students use the definition of the derivative to find $f'(5)$ for $f(x) = \sqrt{3x + 1}$ , as shown. Which solution is right, and why?

\begin{aligned} &\textbf{Solution A} \\ \text{A1:}\quad & \frac{f(5 + h) - f(5)}{h} = \frac{\sqrt{16 + 3h} - 4}{h} \\[4pt] \text{A2:}\quad & = \frac{(16 + 3h) - 16}{h\bigl(\sqrt{16 + 3h} + 4\bigr)} = \frac{3}{\sqrt{16 + 3h} + 4} \\[4pt] \text{A3:}\quad & \to \frac{3}{4 + 4} = \frac{3}{8} \ \text{ as } h \to 0 \\[14pt] &\textbf{Solution B} \\ \text{B1:}\quad & \frac{f(5 + h) - f(5)}{h} = \frac{\sqrt{16 + 3h} - 4}{h} \\[4pt] \text{B2:}\quad & = \frac{(16 + 3h) - 16}{h\sqrt{16 + 3h} + 4h} = \frac{3h}{h\sqrt{16 + 3h} + 4h} \\[4pt] \text{B3:}\quad & \text{at } h = 0 \text{ this is } \tfrac{0}{0}, \text{ so } f'(5) \text{ does not exist} \end{aligned}

A is right; B puts h = 0 in while h still divides the top and the bottom, so its 0/0 means the work is unfinished.
B is right; A cancels h, but h tends to 0, and you cannot divide the top and bottom by something that tends to 0.
Both are right: the quotient is 0/0 at h = 0, as B says, so f′(5) has no value there, and A’s 3/8 is only the value it approaches.
Neither is right: f(5 + h) is √(16 + h), since h is added to the inside of the root, so both start from the wrong quotient.

How sure are you?

SureNot sureGuessing

Theory UBC Math 100 §3.3 CLP-1 §2.2Key results

Worked solution

Strategy

Check each line against the definition. Both solutions set up the quotient and multiply by the conjugate correctly; they differ in what they do with the factor h.

Working

$f(5 + h) = \sqrt{3(5 + h) + 1} = \sqrt{16 + 3h}, \qquad f(5) = 4$
A1 and B1 are right: x is replaced by 5 + h everywhere.
$\frac{\sqrt{16 + 3h} - 4}{h}\cdot\frac{\sqrt{16 + 3h} + 4}{\sqrt{16 + 3h} + 4} = \frac{3h}{h\bigl(\sqrt{16 + 3h} + 4\bigr)}$
Multiplying by the conjugate clears the root from the top. A2 and B2 both reach this quotient; B has multiplied out the bottom.
$\frac{3h}{h\sqrt{16 + 3h} + 4h} = \frac{3h}{h\bigl(\sqrt{16 + 3h} + 4\bigr)} = \frac{3}{\sqrt{16 + 3h} + 4}, \quad h \neq 0$
The bottom of B2 still has the factor h. Cancel it, as A2 does; h is never 0 in the limit, so this is allowed.
$\lim_{h \to 0} \frac{3}{\sqrt{16 + 3h} + 4} = \frac{3}{8}$
Now h = 0 can be put in, as in A3. B3 put it in one step too early and read the resulting 0/0 as “does not exist”.

Answer

\text{A is right:}\quad f'(5) = \frac{3}{8}

Check

At h = 0.01 the quotient (√16.03 − 4)/0.01 is about 0.3748, close to 3/8 = 0.375, so the limit exists.

Where this goes wrong

Reading 0/0 at h = 0 as “no derivative”. In a difference quotient the top and the bottom always both tend to 0; the factor h has to cancel before h = 0 can be put in.

Leans on

Rationalising with the conjugate, from Limits

Try a similar question

Let

f(x) = \sqrt{10 - 3x}

. Use the definition of the derivative to find

f'(2)

, and then find an equation of the tangent line to

y = f(x)

x = 2

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The definition of the derivative practice question 24