Practice: The Derivative

Recognition · Error diagnosis

$f (x) = | x |$ has no derivative at $x = 0$ .

Which statement gives the reason?

2 hints available, least help first.

Hint 1: Retrieval cue

Compute $\frac{| h |}{h}$ for $h = 0.1$ and for $h = - 0.1$ .

Hint 2: Concept cue

One side gives $+ 1$ and the other $- 1$ . What does a two-sided limit require?

Direct application

Use the limit definition to find $f^{'} (5)$ for $f (x) = x^{2}$ .

Simplify $\frac{f (5 + h) - f (5)}{h}$ before letting $h \to 0$ , and give the value of the limit.

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

Expand $(5 + h)^{2}$ and subtract 25. What is left in the numerator?

Hint 2: Concept cue

The numerator is $10 h + h^{2}$ , and every term carries a factor of $h$ to cancel.

Direct application

Let $f (x) = x^{- 2}$ .

Compute $f^{'} (4)$ , giving your answer as a decimal.

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

Apply $n x^{n - 1}$ with $n = - 2$ . What is $n - 1$ ?

Hint 2: Concept cue

$f^{'} (x) = - 2 x^{- 3}$ , and $4^{3} = 64$ .

Direct application

Let $f (x) = x^{2}$ and $g (x) = x^{3}$ .

Use the product rule to compute $(f g)^{'} (2)$ .

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

The rule has two terms. What are $f^{'} (2)$ , $g (2)$ , $f (2)$ and $g^{'} (2)$ ?

Hint 2: Concept cue

Simplify $x^{2} \cdot x^{3}$ to $x^{5}$ and differentiate that as a check on your answer.

Direct application

Let $y = (3 x^{2} + 1)^{4}$ .

Compute $y^{'} (1)$ .

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

What is the outer function, and what is the inner one?

Hint 2: Concept cue

After differentiating the fourth power, multiply by the derivative of $3 x^{2} + 1$ .

Construction · Direct application · Explanation

(a) Use the limit definition to find $f^{'} (a)$ for $f (x) = x^{2} - 4 x$ , showing the cancellation before the limit is taken. Confirm your formula at $a = 3$ against a numerical difference quotient.

(b) Differentiate $y = \frac{x^{2} + 1}{x - 1}$ and evaluate $y^{'} (3)$ .

(c) Differentiate $y = (2 x^{3} - 5)^{4}$ and evaluate $y^{'} (1)$ .

(d) For $f (x) = x^{3} - 3 x$ , find every critical point, classify each, and state whether either is a global extremum.

(e) A learner writes: " $f^{'} (c) = 0$ , so $f$ has a minimum or a maximum at $c$ ." Explain what is wrong, give a counterexample, and say what additional information would settle the question.

Write your answer

Write your answer, then compare it with the worked solution.

3 hints available, least help first.

Hint 1: Retrieval cue

In (a), expand fully and collect terms before dividing. Every surviving term should carry a factor of $h$ .

Hint 2: Concept cue

For (d), solve $f^{'} (x) = 0$ first, then classify with $f^{″}$ or with the sign of $f^{'}$ on either side.

Hint 3: Strategy cue

In (e), look for a function whose derivative vanishes at a point where it is still strictly increasing.

Compare with the worked solution

Comparing does not record a result. Judging your own written answer cannot show that you can do this without help.

(a) From the definition.

\frac{f (a + h) - f (a)}{h} = \frac{[(a + h)^{2} - 4 (a + h)] - [a^{2} - 4 a]}{h} .

Expanding the numerator: $a^{2} + 2 a h + h^{2} - 4 a - 4 h - a^{2} + 4 a = 2 a h + h^{2} - 4 h$ . Every term carries a factor of $h$ , so

\frac{2 a h + h^{2} - 4 h}{h} = 2 a + h - 4,

and letting $h \to 0$ gives $f^{'} (a) = 2 a - 4$ . The cancellation must precede the limit: at $h = 0$ the unsimplified quotient is $0 / 0$ . Check at $a = 3$ . The formula gives $f^{'} (3) = 2$ . Numerically with $h = 0.001$ : $f (3.001) = {3.001}^{2} - 4 (3.001) = 9.006001 - 12.004 = - 2.997999$ and $f (3) = 9 - 12 = - 3$ , so the quotient is $\frac{0.002001}{0.001} = 2.001$ , matching $2 a + h - 4 = 2.001$ exactly ✓. (b) Quotient rule. With $u = x^{2} + 1$ , $u^{'} = 2 x$ , $v = x - 1$ , $v^{'} = 1$ :

y^{'} = \frac{u^{'} v - u v^{'}}{v^{2}} = \frac{2 x (x - 1) - (x^{2} + 1)}{(x - 1)^{2}} = \frac{x^{2} - 2 x - 1}{(x - 1)^{2}} .

At $x = 3$ : numerator $= 9 - 6 - 1 = 2$ , denominator $= 4$ , so $y^{'} (3) = \frac{1}{2}$ . The order $u^{'} v - u v^{'}$ matters; reversing it flips the sign. (c) Chain rule. Outer $u^{4}$ , inner $u = 2 x^{3} - 5$ with $u^{'} = 6 x^{2}$ :

y^{'} = 4 (2 x^{3} - 5)^{3} \cdot 6 x^{2} = 24 x^{2} (2 x^{3} - 5)^{3} .

At $x = 1$ : $u = 2 - 5 = - 3$ , so $y^{'} = 24 \cdot 1 \cdot (- 27) = - 648$ . The negative sign comes from the odd power of a negative inner value, and omitting $u^{'}$ would give $4 (- 27) = - 108$ . (d) Critical points of $f (x) = x^{3} - 3 x$ .

f^{'} (x) = 3 x^{2} - 3 = 3 (x - 1) (x + 1) = 0 ⟹ x = \pm 1 .

Classify with $f^{″} (x) = 6 x$ : | $x$ | $f (x)$ | $f^{″} (x)$ | verdict |
|---|---|---|---|
| $- 1$ | $2$ | $- 6$ | local maximum |
| $+ 1$ | $- 2$ | $+ 6$ | local minimum | Confirming by sign changes in $f^{'}$ : $f^{'} (- 2) = 9 > 0$ , $f^{'} (0) = - 3 < 0$ , $f^{'} (2) = 9 > 0$ . The function rises, falls, then rises. A maximum at $- 1$ and a minimum at $+ 1$ ✓. Neither is global. $f (3) = 27 - 9 = 18$ , which exceeds the local maximum value of 2, and $f (- 3) = - 27 + 9 = - 18$ , below the local minimum of $- 2$ . As $x \to \pm \infty$ the cubic term dominates and $f$ is unbounded in both directions, so no global extremum exists. (e) What a zero derivative establishes. The claim is false: $f^{'} (c) = 0$ is necessary for an interior extremum of a differentiable function, but not sufficient. Counterexample. $f (x) = x^{3}$ has $f^{'} (x) = 3 x^{2}$ , so $f^{'} (0) = 0$ . But $x^{3}$ is strictly increasing everywhere, for any $ε > 0$ , $f (- ε) = - ε^{3} < 0 < ε^{3} = f (ε)$ , so 0 is neither a maximum nor a minimum. It is an inflection point with a horizontal tangent. What would decide it. Either: - The sign of $f^{'}$ on both sides of $c$ . A change from positive to negative gives a maximum, negative to positive a minimum, and no change gives neither. This test always works.
- The sign of $f^{″} (c)$ , when it is nonzero. Negative gives a maximum, positive a minimum. The second test is silent when $f^{″} (c) = 0$ , and that silence is genuine rather than a gap in technique: at $x = 0$ the functions $x^{3}$ , $x^{4}$ and $- x^{4}$ all have $f^{'} = f^{″} = 0$ , and have respectively no extremum, a minimum and a maximum. Only the first-derivative sign test distinguishes them. The general point. A derivative reports local behaviour at a point. Deciding whether a critical point is an extremum requires information about a neighbourhood, and deciding whether it is global requires information about the whole domain, including endpoints and limiting behaviour, neither of which any derivative at $c$ can supply.

A complete answer does each of these:

evaluates from the limit
applies power rule
applies product rule
applies chain rule
diagnoses non differentiability
interprets the derivative

Practice: The Derivative

Recognition · Error diagnosis

Direct application

Direct application

Direct application

Direct application

Construction · Direct application · Explanation

Session complete

Settings

Focus timer

Today

Last seven days

Where the time went

Recent sessions

Settings

Notifications

Your history