OMPT drill — Calculus

Differentiation rules

Power, product, quotient, and chain rule. Chain rule with a function inside a function inside a power is the OMPT's favourite way to separate the prepared from the hopeful. Say the rule you are using out loud before you differentiate or integrate — it sounds silly and it works.

Tested inOMPT-A OMPT-B OMPT-D OMPT-F

Lesson

The derivative $f' (x)$ measures how fast $f$ is changing at each point: the slope of the graph, read as a function of position. Computing it from the definition every time would be misery, so calculus hands you a rulebook instead, and this topic is that rulebook. The foundation is the power rule: $x^{n}$ differentiates to $n x^{n - 1}$ . Bring the exponent down as a multiplier, knock the exponent down by one. $x^{5} \to 5 x^{4}$ .

Three combination rules then cover every function the OMPT can build: the product rule for things multiplied, the quotient rule for things divided, and the chain rule for things nested inside each other. Of the three, the chain rule decides grades. The exam’s favourite construction is a function inside a function inside a power, precisely because it forces you to apply the chain rule deliberately rather than on autopilot. My advice for the whole topic: before differentiating anything, say out loud which rule the expression needs. The five seconds of diagnosis prevent most of the errors.

Problem 1Numeric answer

Let $f (x) = x^{3} - 5 x^{2} + 2 x$ . Compute $f^{'} (2)$ .

Show the worked solution

Answer: -6

The power rule gives $f^{'} (x) = 3 x^{2} - 10 x + 2$ . Evaluating at $x = 2$ : $3 \cdot 4 - 20 + 2 = 12 - 20 + 2 = - 6$ . The curve is falling at that point.

Problem 2Multiple choice

What is the derivative of $y = x^{2} e^{x}$ ?

A $2 x e^{x}$
B $(x^{2} + 2 x) e^{x}$
C $2 x + e^{x}$
D $x^{2} e^{x - 1}$

Show the worked solution

Answer: B — $(x^{2} + 2 x) e^{x}$

This is a product, so I need the product rule: $(x^{2})^{'} e^{x} + x^{2} (e^{x})^{'} = 2 x e^{x} + x^{2} e^{x}$ . Factoring out $e^{x}$ gives $(x^{2} + 2 x) e^{x}$ .

Problem 3Numeric answer

Let $g (x) = (3 x + 1)^{5}$ . Compute $g^{'} (0)$ .

Show the worked solution

Answer: 15

The chain rule gives $g^{'} (x) = 5 (3 x + 1)^{4} \cdot 3 = 15 (3 x + 1)^{4}$ . At $x = 0$ the bracket equals 1, so $g^{'} (0) = 15$ .

Problem 4Multiple choice

What is the derivative of $y = ln (x^{2} + 1)$ ?

A $\frac{1}{x ^{2} + 1}$
B $\frac{2 x}{x ^{2} + 1}$
C $\frac{1}{2 x}$
D $2 x ln (x^{2} + 1)$

Show the worked solution

Answer: B — $\frac{2 x}{x ^{2} + 1}$

The chain rule for logarithms says the derivative is the inner derivative over the inner function: $\frac{( x ^{2} + 1 ) ^{'}}{x ^{2} + 1} = \frac{2 x}{x ^{2} + 1}$ .

Problem 5Spot the error

A student differentiates $h (x) = \frac{x}{x + 1}$ . Step 1: identify numerator $u = x$ and denominator $v = x + 1$ , with $u^{'} = 1$ and $v^{'} = 1$ . Step 2: apply the quotient rule as $h^{'} = \frac{u v ^{'} - v u ^{'}}{v ^{2}}$ . Step 3: substitute to get $h^{'} (x) = \frac{x - ( x + 1 )}{( x + 1 ) ^{2}} = \frac{- 1}{( x + 1 ) ^{2}}$ . Which step contains the error?

Show the worked solution

Answer: Step 2

The quotient rule is $h^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ — Step 2 has the numerator terms swapped, which flips the sign of the whole result. Done correctly: $\frac{1 \cdot ( x + 1 ) - x \cdot 1}{( x + 1 ) ^{2}} = \frac{1}{( x + 1 ) ^{2}}$ . A quick sanity check: $\frac{x}{x + 1}$ is increasing for $x > - 1$ , so its derivative must be positive there.

Problem 6Numeric answer

Let $p (x) = sin x cos x$ . Compute $p^{'} (\frac{π}{3})$ as a decimal.

Show the worked solution

Answer: -0.5

By the product rule, $p^{'} (x) = cos^{2} x - sin^{2} x$ , which is exactly $cos 2 x$ . At $x = \frac{π}{3}$ : $cos \frac{2 π}{3} = - \frac{1}{2}$ . Spotting that $sin x cos x = \frac{1}{2} sin 2 x$ gets there even faster.