CEGEP Calculus I — The Derivative Worksheet

All 6 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. Q1Average and Instantaneous Rate of Change

   A construction hoist climbs the outside of a tower. Its height above the loading platform, in metres, is $$h(t)=\frac{1}{2}{t}^2+3t,0\le t\le 10,$$ where $t$ is measured in seconds.

   1. Find the average rate of change of $h$ on the interval $[2,6]$, with units.
   2. Use the limit ${\displaystyle {lim}_{k\to 0}\frac{h(2+k)-h(2)}{k}}$ to find the instantaneous rate of change of $h$ at $t=2$, with units.
   3. The two answers are different. Which one is the reading on the hoist's speedometer at the instant $t=2$, and why?
2. Q2The Definition of the Derivative

   Let $f(x)=\sqrt{3x+4}$.

   1. Use the definition ${\displaystyle f^{\prime }(x)={lim}_{h\to 0}\frac{f(x+h)-f(x)}{h}}$ to find $f^{\prime }(x)$.
   2. State the domain of $f$ and the domain of $f^{\prime }$ in interval notation, and say why they are not the same.
3. Q3The Derivative as a Function

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

   1. Use the definition of the derivative to find the function $f^{\prime }$.
   2. State the domain of $f^{\prime }$, and sketch $y=f^{\prime }(x)$ on the grid below.
   3. Using your sketch, state the interval on which the values of $f$ are falling as $x$ increases, and the values of $x$ at which $f$ has a horizontal tangent. Justify each answer by referring to the sign or the value of $f^{\prime }$.

   A blank Cartesian grid for this question is on the printable PDF.

4. Q4Differentiability and Continuity

   Let $f(x)=|x^2-9|$.

   1. Show that $f$ is continuous at $x=3$.
   2. Compute ${\displaystyle {lim}_{x\to 3^{-}}\frac{f(x)-f(3)}{x-3}}$ and ${\displaystyle {lim}_{x\to 3^{+}}\frac{f(x)-f(3)}{x-3}}$, and deduce whether $f^{\prime }(3)$ exists.
   3. What do (a) and (b) together show about the relationship between continuity and differentiability?
5. Q5The Equation of the Tangent Line

   Let $f(x)={\displaystyle \frac{4}{x}}$.

   1. Use the definition of the derivative to show that $f^{\prime }(x)=-{\displaystyle \frac{4}{x^2}}$ for $x\ne 0$.
   2. Find the equation of the tangent line to the graph of $f$ at the point where $x=2$.
   3. Verify algebraically that this tangent meets the graph of $f$ at that point only.
6. Q6Synthesis — drawing on several topics in this set

   A reservoir is drawn down over a dry summer. Its volume, in millions of cubic metres, is $$V(t)=\frac{1}{2}(12-t{)}^2,0\le t\le 12,$$ where $t$ is the number of days since drawdown began.

   1. Find the average rate of change of $V$ on $[0,4]$, with units, and say what its sign means.
   2. Use the definition of the derivative to find $V^{\prime }(t)$, and evaluate $V^{\prime }(4)$ with units.
   3. Write the equation of the tangent line to the graph of $V$ at $t=4$, and use it to estimate $V(5)$.
   4. Compare your estimate with the exact value of $V(5)$, and say whether the tangent line over-estimates or under-estimates. Then state the day on which the tangent line predicts the reservoir is empty, and the day it is actually empty.

The 4 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.

An average over an interval is not a rate at an instant

Q1 puts the two side by side on purpose. You compute the average rate of change of a hoist's height over [2, 6], then the instantaneous rate at t = 2 from a limit, and the two numbers disagree. Part (c) asks which one the speedometer shows, and the answer is not a matter of taste: an average uses only the two endpoint heights and says nothing whatever about any single moment in between.

Notice that the instantaneous rate is not one of the averages. Every interval of positive width gives an average; shrinking the width gives a sequence of them; the derivative is what they approach and is generally equal to none of them. That gap is the whole reason the definition needs a limit rather than a smaller interval.

The difference quotient, and the two shapes it is written in

These are the same statement. Put x = a + h and the second becomes the first term for term, with h → 0 saying exactly what x → a says. Which one you use is a question of convenience: the h-form suits an algebraic expansion, the x-form suits an expression that already factors. Recognising both matters, because a limit handed to you in either shape is a derivative in disguise, and naming the f and the a is often faster than evaluating it.

Where f and f′ stop agreeing about the domain

Q2 asks for both domains in interval notation and then for the reason they differ. It is worth taking seriously, because the answer is not a technicality. A square root is defined at the endpoint where its radicand is zero, so f is happy there; the difference quotient at that endpoint has a root of h over h, which grows without bound as h approaches 0 from the right. The function has a point; the graph has no finite slope at it.

From a slope at one point to a function

Q3 is the pivot of the set. Up to that point a derivative has been one number attached to one input; here you carry the x through the whole computation and come out with a rule. That rule is a new function whose input is a position and whose output is the slope of the original graph there — so the graph of f′ is a picture of every slope of f at once.

That is what part (c) is testing when it asks you to read intervals off your sketch. Where f′ is negative, the values of f are falling; where f′ is zero, the tangent to f is horizontal. Justify by naming the sign or the value of f′ rather than by pointing at the picture — "f′ < 0 on this interval, so f decreases there" is the sentence, and it is the sentence the marker is looking for. Note the direction of the reasoning: you are reading f from f′, not the other way round.

Differentiability implies continuity — and never the reverse

Q4 hands you a function that is continuous at a point and asks whether it has a derivative there. Part (a) settles continuity; part (b) computes the two one-sided difference quotients separately; part (c) asks what the pair of results establishes.

The technique matters as much as the conclusion. When a rule changes at a point — an absolute value, a piecewise definition — you cannot compute one difference quotient and take its limit, because there is no single expression to take it of. You compute the left-hand and right-hand limits from their own branches and compare them, exactly as you would for any two-sided limit.

A tangent line is a point and a slope, and nothing else

Q5 asks for the tangent to a graph at a given input. The recipe never changes: evaluate the function there to get the point, evaluate the derivative there to get the slope, then write the point-slope form and tidy it. The derivative supplies one of the two ingredients, which is why forgetting to compute the y-coordinate is the most common way to lose the mark.

The tangent line as a local approximation

Q6 is the synthesis question and it uses nearly everything above: an average rate of change with its sign interpreted, a derivative from the definition, a tangent line, and then the tangent line used to predict a value the function actually takes. Comparing the prediction with the exact value is the part that carries the idea.

Getting the most out of it

Do not reach for a rule you have not been given

Everything on this sheet is done from the difference quotient. If you already know the power rule, use it to check your answer and not to produce it — a question that says "using the definition" awards the marks for the limit, and a correct derivative with no difference quotient above it scores close to nothing. The rules arrive in the next worksheet of this course, and they are much easier to trust once you have seen where they come from.

Expand fully before you subtract

Write f(x + h) out in full on its own line first. Most of the algebra in this set fails at the expansion, not at the limit, and the failure is invisible later: once a term that should have cancelled survives into the numerator, the h will not factor out and the quotient goes nowhere.

Say "for h ≠ 0" every time you cancel

One clause, written as you divide. It is the reason the cancellation is allowed, it costs nothing, and its absence is what turns an otherwise correct page into an argument that appears to divide by zero at the decisive step.

Carry the units, and say what the sign means

These questions are set in metres, seconds, days and cubic metres because a rate of change means something concrete: metres per second, millions of cubic metres per day. A negative derivative is a quantity that is falling. Marks are attached to that sentence in the ministerial competency's performance criteria, and it is the one students skip.

Common questions