CEGEP Calculus I — Differentiation Rules Worksheet

All 7 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Basic Differentiation Formulas

   Differentiate each function. Rewrite first where that avoids a rule you have not needed yet, and state any restriction on the domain.

   1. $f(x)=5x^4-3x^2+7x-2$
   2. $g(x)={\displaystyle \frac{4}{x^3}}+\sqrt[3]{x}$
   3. $h(x)={\displaystyle \frac{2x^3-x+6}{x^2}}$
2. Q2The Product Rule

   Let $f(x)=(3x^2-5)(x^3+2x)$.

   1. Find $f^{\prime }(x)$ using the product rule, and simplify.
   2. Expand $f(x)$ first and differentiate the resulting polynomial. Confirm that the two answers agree.
3. Q3The Quotient Rule

   Let $f(x)={\displaystyle \frac{2x-1}{x^2+3}}$.

   1. Find $f^{\prime }(x)$ and simplify the numerator.
   2. Find the equation of the tangent line to $y=f(x)$ at $x=1$.
4. Q4The Chain Rule

   Differentiate, simplifying where a common factor allows it.

   1. $f(x)=(2x^3-5x{)}^4$
   2. $g(x)=\sqrt{9-x^2}$ (state the set on which $g^{\prime }$ exists)
5. Q5Implicit Differentiation

   The curve $C$ has equation $x^2+3xy-y^2=9$.

   1. Verify that the point $(3,0)$ lies on $C$.
   2. Find ${\displaystyle \frac{dy}{dx}}$ in terms of $x$ and $y$.
   3. Find the equation of the tangent to $C$ at $(3,0)$.
6. Q6Higher-Order Derivatives

   An ore trolley runs on a straight track. Its position, in metres from a marker on the track, is modelled by $s(t)=t^3-6t^2+9t$ for $0\le t\le 4$, with $t$ in seconds.

   1. Find the velocity $v(t)=s^{\prime }(t)$ and the acceleration $a(t)=s^{″}(t)$, with units.
   2. Find the time at which the acceleration is zero, and state what is happening to the velocity at that instant.
   3. A student says “the acceleration is zero at that time, so the trolley is momentarily at rest”. Explain why this is wrong, and give the times in $[0,4]$ at which the trolley really is at rest.
7. Q7Synthesis — drawing on several topics in this set

   A directional antenna sits at the top of a mast, $3$ m above level ground. A receiver placed on the ground at a horizontal distance of $d$ metres from the foot of the mast reports a normalised line-of-sight reading — the horizontal separation divided by the straight-line separation — modelled by $$R(d)=\frac{d}{\sqrt{d^2+9}},d\ge 0.$$

   1. Find $R^{\prime }(d)$ using the quotient rule together with the chain rule, and simplify to a single term.
   2. Evaluate $R^{\prime }(4)$ to three decimal places, with units.
   3. Obtain $R^{\prime }(d)$ a second way, by writing $R(d)=d(d^2+9{)}^{-1/2}$ and using the product rule. Then use the sign of $R^{\prime }$ to describe how the reading behaves as $d$ increases.

The 6 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

Why the early questions look easier than your textbook's

The sheets are ordered so that nothing before The Chain Rule needs a composition, and nothing before The Quotient Rule needs a quotient. That is deliberate, and it is worth knowing, because a textbook chapter is usually written the other way round: every exercise set assumes every rule in the chapter. Here Q1 can be done with the power rule alone, Q2 needs the product rule and nothing after it, and so on down the page.

Q1 — rewrite first, then differentiate

The power rule differentiates xⁿ to n·xⁿ⁻¹ for every real n, not only for whole numbers — but only once the expression is actually written as a power. Q1 gives you three functions that are each one algebraic step away from being a sum of powers.

Q2 — the product rule is not the product of the derivatives

Worth saying flatly, because it is the wrong instinct almost everyone has once. The derivative of a product has two terms:

Q2 is built so you can prove this to yourself rather than take it on trust. Part (a) applies the rule; part (b) expands the product into an ordinary polynomial and differentiates that instead. Two routes, one function, so the answers must agree — and if you also compute u′v′ you can watch it fail to match either of them. That is a more convincing demonstration than any assertion on this page.

Q3 — the quotient rule, and the order that is not symmetric

Swapping the two products in the product rule changes nothing, because addition commutes. Swapping them in the quotient rule negates your entire answer, because subtraction does not.

Q3 then asks for a tangent line, which needs two numbers and not one: the slope f′(1) and the point f(1). Write it in point-slope form, y − f(a) = f′(a)(x − a), before simplifying. Two habits pay off here. Simplify the numerator and leave the denominator factored — later questions ask where f′ = 0, and only the numerator can answer that, since a fraction vanishes only where its top does. And sanity-check the sign of your slope against the function itself: evaluate f a little to each side of the point, see whether it is rising or falling, and compare.

Q4 — the chain rule, and where the derivative stops existing

A composite is differentiated from the outside in: the derivative of the outer function, evaluated at the inner function left untouched, multiplied by the derivative of the inner function.

Q4(b) attaches a second question to the derivative: state the set on which g′ exists. This is not decoration. Differentiating a square root sends the inside into a denominator, so the derivative can fail to exist at points where the function itself is perfectly well defined — the endpoints of a closed domain are the usual casualties. The domain of f′ is a subset of the domain of f, never the other way round, and it is not automatically the same set. Give it in standard interval notation.

Q5 — implicit differentiation, and what dy⁄dx is doing there

When a curve is given by an equation in x and y that you cannot or would rather not solve for y, you differentiate the whole equation with respect to x and treat y as a function of x throughout. Every y that gets differentiated leaves a dy⁄dx behind; that is the chain rule, applied to a function whose formula you never wrote down.

Do not solve for y first, even when you could. On the curve in Q5 that would mean the quadratic formula, a square root, and then a chain rule on that root — three chances to go wrong in place of one line of algebra. Implicit differentiation is the shorter route even when the explicit one exists.

Q6 — higher-order derivatives, and the sentence people get backwards

Differentiating again gives the second derivative, written f″(x) or d²y⁄dx². In motion the chain is short, and worth holding as words rather than symbols: position, then velocity, then acceleration.

Units carry down the chain as well, and Q6(a) asks for them. If position is in metres and time in seconds, velocity is metres per second and acceleration metres per second squared. Each differentiation divides by one more unit of the input variable — a rule that keeps working long after this question, in every rate you interpret.

Q7 — when several rules meet in one function

The closing question is the realistic one: a quotient whose denominator is a composite, so the chain rule runs inside the quotient rule. Nothing new is being tested, only whether you can keep two rules straight at the same time.

Q7 also asks what the sign of R′ says about the reading as the distance grows. That is the step students skip, and it is where the marks for interpretation live. A derivative positive throughout the domain means the quantity is strictly increasing — a sentence about the physical situation, not about the algebra. The numerical value is wanted to three decimal places and with units; a bare number answers a different question from the one asked.

Getting the most out of it

Choose the rule before you compute anything

Look at the function as a whole first and name its outermost structure: is it a sum, a product, a quotient, or a composite? That name decides the first line of your work. Most lost marks in this set are not arithmetic; they are a product rule applied to a composite, or a quotient rule applied to something that was a sum of powers all along.

Rewrite anything that will simplify

Fractional and negative exponents, cancelling a common factor, dividing through by a single term in the denominator — each of these can remove a whole rule from the calculation. A step you never take is a step you cannot get wrong.

Check by a second route

A product of polynomials can be expanded and differentiated instead. A quotient can be rewritten as a product with a negative exponent. An implicit result can be tested at a known point on the curve. Q2 and Q7 build the check into the question; do it on the others too, and the sign errors that are otherwise invisible stop reaching your final answer.

Answer the whole question

Domain restrictions, units, interval notation, the sentence of interpretation, the reason a claim is false — these are separate marks from the derivative itself, and they are the cheapest ones on any paper. They are also the ones most often left blank.

Common questions