CEGEP Calculus I Derivatives Free · no sign-up

CEGEP Calculus I — Derivatives of Transcendental Functions

By Marius — private math tutor in Montreal. Updated 30 August 2026

Every function on this sheet is transcendental: exponentials in base e and in any other base, logarithms in any base, the trigonometric functions, the inverse trigonometric ones, and the expressions where the sensible move is to take a logarithm before differentiating at all. The rules themselves are one line each and quickly learned. What the questions keep coming back to is the domain — on which set your derivative is actually valid, and why that set is so often smaller than the one you started with. Read the questions on this page, then print the free PDF when you want room to write.

Practice worksheet — free PDF

4 pages 6 questions Letter size, print-ready

Download the worksheet Read the questions here

No email, no account, no watermark. Teachers: photocopy it for your classes freely. Worked solutions and 4 harder problems come with the CEGEP Calculus I bundle.

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.

Q1Derivatives of Exponential Functions

Differentiate each function. Factor your answer where a common factor appears.
1. $f (x) = x^{3} e^{2 x}$
2. $g (t) = 5^{t^{2} - t}$
3. $h (x) = \frac{e^{x}}{1 + e^{x}}$
Q2Derivatives of Logarithmic Functions

Differentiate each function and state the domain on which your derivative is valid.
1. $f (x) = \ln (3 x^{2} + 5)$
2. $g (x) = x^{2} \ln (4 x)$
3. $h (x) = \log_{2} (x^{3} + 1)$
Q3Derivatives of Trigonometric Functions

Differentiate and simplify as far as possible.
1. $f (x) = x \sin x + \cos x$
2. $g (x) = \tan (3 x^{2})$
3. $h (x) = \frac{\cos x}{1 - \sin x}$
Q4Derivatives of Inverse Trigonometric Functions

Differentiate each function.
1. $f (x) = \arctan (3 x)$
2. $g (x) = \arcsin (x^{2})$ ; state the interval on which $g^{'}$ exists.
3. $h (x) = x \arctan x - \frac{1}{2} \ln (1 + x^{2})$ ; simplify completely and comment on what you obtain.
Q5Logarithmic Differentiation

Use logarithmic differentiation to find $\frac{d y}{d x}$ .
1. $y = x^{\cos x}$ , for $x > 0$ .
2. $y = \frac{(x + 2)^{4} \sqrt{2 x - 1}}{{(x^{2} + 3)}^{5}}$ , for $x > \frac{1}{2}$ .
In each case, say which feature of the expression makes logarithmic differentiation the right tool.
Q6Synthesis — drawing on several topics in this set

A damped oscillation is described by $f (x) = e^{- x} \sin (2 x)$ .
1. Find $f^{'} (x)$ and write it with $e^{- x}$ factored out.
2. Find the equation of the tangent line to the curve $y = f (x)$ at $x = 0$ .
3. Show that $f^{'} (x) = 0$ if and only if $\tan (2 x) = 2$ . Give the smallest positive solution exactly, using $\arctan$ , and then to three decimals.

The 4 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

Which stream is this for? This is CEGEP Calculus I — 201-NYA-05 under the legacy numbering, 201-SN2-RE under the current one, the differential calculus course of the Sciences de la nature programme. It is written against competency 0M02, formerly 00UN. The set follows Differentiation Rules and assumes it: the product, quotient and chain rules are used on nearly every question rather than taught here, and implicit differentiation is what makes logarithmic differentiation work. What is new is the derivative of each transcendental function. One thing to note about scope — there is no integration anywhere on this sheet. Antiderivatives and the substitution rule sit under competency 0M03, which is Calculus II, so if that is what you came for it is the next course rather than a missing section here. Indeterminate forms and L'Hospital's rule come after this set, in the same course.

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.

Which of these families is actually new

The ministerial devis draws a line through this material that is worth knowing about, because it tells you where to spend the time. Exponential, logarithmic and trigonometric functions — their graphs, the exponent laws, the properties of logarithms, the trigonometric circle and the standard identities — are listed as concepts you arrive with. The inverse trigonometric functions are listed the other way, among the concepts to be acquired in this course, and two are named outright: arcsine and arctangent.

So if the sheet feels uneven, that is deliberate rather than accidental. Q1, Q2 and Q3 attach new derivative rules to functions you already know how to draw. Q4 is the one where the function itself is unfamiliar, and it is the one to slow down on — you cannot sanity-check an answer against a graph you have never pictured.

Exponentials: base e is the easy case, and it is not the general one

Two rules, and it is the second that goes missing under time pressure. In base e the derivative reproduces itself; in any other base a constant factor ln a appears and does not go away.

(e^u)′ = e^u·u′ (a^u)′ = a^u ln a·u′ (a > 0, a ≠ 1)

Q1 puts the three rules you already own to work on exponentials: a product in part a), a general base in part b), a quotient in part c). Part b) separates the people who learned the rule from the people who learned the special case. And the instruction to factor your answer in part a) is not decoration — both terms carry a common exponential factor, and a factored derivative is one whose sign and whose zeros you can read straight off.

A free check on any exponential derivative: e^u and a^u are strictly positive for every real input, never zero and never negative. So the sign of a derivative is decided entirely by whatever multiplies the exponential. If your working produces something that changes sign where no other factor does, the error is in the working, not in the function.

The domain of ln comes before the derivative, not after it

Q2 asks you to differentiate and state the domain on which your derivative is valid, and those are two separate marks. The rules are short:

(ln u)′ = u′ ⁄ u (log_a u)′ = u′ ⁄ (u ln a)

The trap is that u′⁄u is defined in places where ln u never was. A logarithm needs a strictly positive argument, so before you differentiate anything, solve u > 0. That set is the domain of the function, and the derivative cannot be valid outside it however healthy the quotient looks. Solve the inequality rather than eyeballing it: the three arguments in Q2 are of three different shapes — a quadratic, a scalar multiple of x, a cubic — and each needs its own two lines of work. Only one of the three is settled at a glance, and it is not the one most people skip.

A related habit worth building: apply the logarithm laws before you differentiate, whenever the argument is a product, a quotient or a power. A constant inside a logarithm splits off as an additive constant and then differentiates to nothing — which is why a factor that looks like it must survive into the answer sometimes does not appear there at all.

Two notation points, both house rules and both marked. Write ln x for the natural logarithm, never log_e x. And give domains in standard interval notation — (0, ∞), (−1, ∞), (−∞, ∞) — with a round bracket at every infinite end and at every value the function does not reach.

Why ln(x²) is 2 ln|x| and not 2 ln x

This is the most common invalid step in the whole topic, and it is invalid for a reason worth stating precisely rather than memorising.

ln(x²) = 2 ln|x| for every x ≠ 0 ln(x²) = 2 ln x only where x > 0

The left-hand side is defined for every x except 0, because squaring makes the argument positive whichever side you came from. The right-hand side, 2 ln x, is defined only for x > 0. Two expressions with different domains are not equal, so the identity that holds in general is the one carrying the absolute value.

Where a question has already restricted you to positive x the absolute value is harmless and you may drop it — but drop it because you checked the domain, not by reflex. The same caution applies to ln(uv) = ln u + ln v: that split needs u and v individually positive, which is a stronger condition than their product being positive.

Trigonometric derivatives, and the identity that finishes the job

The derivatives themselves are worth knowing cold, and the minus signs on cosine, cotangent and cosecant are where marks are lost. Q3 asks you to differentiate and simplify as far as possible, which is the instruction that makes it more than a rules exercise.

(sin u)′ = cos u·u′ (cos u)′ = −sin u·u′ (tan u)′ = sec²u·u′

In part a) the product rule produces two terms, and the sign you give (cos x)′ decides whether the expression tidies up at all; if yours refuses to simplify, check that sign before checking anything else. In part c) the quotient rule leaves a numerator containing sin²x + cos²x, and the Pythagorean identity is what collapses it; a numerator left in that state has not been simplified as far as possible. Then read the restriction off the denominator of the original function. Because sine repeats, that exclusion is a whole family of values, written with an integer k, not a single number.

Part b) is a reminder that the chain rule never goes away in this set. The angle is not x, so the derivative of the angle is a factor of the answer, and writing sec² of the angle without it is the single most expensive slip on the sheet.

arcsin and arctan: the ranges first, then the derivatives

These are the unfamiliar ones, and nearly everything that goes wrong with them follows from not having the domain and the range in front of you. Sine and tangent are not one-to-one, so each inverse is built by restricting the original function first — and that restriction is the range of the inverse.

What the two inverses actually are

Domain, range, and the derivative that follows. The derivative's domain is the line to read twice.

1
arcsin x — domain [−1, 1], range [−π⁄2, π⁄2]
It answers "which angle in that closed range has this sine?". An input outside [−1, 1] is meaningless, and the output never lands in the second or third quadrant however the question is phrased.
2
arctan x — domain (−∞, ∞), range (−π⁄2, π⁄2)
Every real number has an arctangent, and the range brackets are open: ±π⁄2 are approached and never attained, which is the same fact as tangent having vertical asymptotes there.
3
(arcsin u)′ = u′ ⁄ √(1 − u²)
Look at where this can be evaluated. The square root sits in a denominator, so it needs 1 − u² > 0 strictly: the endpoints belong to the function and not to its derivative.
4
(arctan u)′ = u′ ⁄ (1 + u²)
That denominator is at least 1 for every real u, so nothing is ever excluded and no square root appears. It is also the analytic reason arctan never turns back on itself.

Q4 b) asks for the interval on which the derivative exists, and it is not the interval on which the function exists. That gap is the whole point of the part. Work it in two steps: first the condition that makes the arcsine defined — its argument must lie in [−1, 1] — then tighten to the strict inequality the square root in the denominator demands. The endpoints where the derivative dies are exactly where the tangent to the curve is vertical, which you can see on the graph of arcsin at ±1.

Part c) combines an inverse tangent with a logarithm and asks you to simplify completely and comment on what you obtain. Take both instructions seriously: the comment is a mark of its own, and "completely" is doing real work — differentiating term by term gives you an expression that is not yet in its final state, and the tidying is the reason the question is on the sheet. Do not stop at the first line of algebra you write down.

Logarithmic differentiation: two situations, one method

Q5 asks you not only to use the method but to say which feature of the expression makes it the right tool. There are two possible answers, and the question gives you one of each.

The variable is in the base and in the exponent. Then no ordinary rule applies at all. The power rule is stated for a fixed exponent; the exponential rule is stated for a fixed base. An expression with x in both places satisfies neither hypothesis, so reaching for either is not an arithmetic slip — it is using a theorem outside its statement.

The expression is a long product, quotient and power stacked together. Here an ordinary derivative exists and is perfectly findable; taking logarithms is simply cheaper. The logarithm laws turn products into sums, quotients into differences and exponents into coefficients, so nested product and quotient rules become a handful of easy terms with nothing left to simplify at the end.

The method is fixed: take ln of both sides, expand with the logarithm laws, differentiate both sides with respect to x — the left-hand side gives (1⁄y)·dy⁄dx by the chain rule, which is where implicit differentiation earns its keep — then multiply through by y and write y back in terms of x. An answer still containing the letter y is not finished. Notice too why both parts of Q5 hand you a restriction on x: ln y only makes sense while y > 0, and saying so is part of doing the method honestly rather than an ornament on it.

The synthesis question, and what "if and only if" is asking for

Q6 is a damped oscillation — an exponential multiplying a sine — and it runs the set's tools together. Part a) needs the product rule with a chain rule inside each factor, and asks for the answer with the exponential factored out. Again that is not cosmetic: a factored derivative is one whose zeros you can read.

Part b) is a tangent line, and a tangent line always needs two numbers, the value and the derivative at the same point. Computing only one of them is the routine way to lose the part.

Part c) says show that f′(x) = 0 if and only if tan(2x) = 2. An "if and only if" claims two conditions have exactly the same solutions, so an argument travelling one way only is incomplete. Two things need saying. First, an exponential factor is never zero, so it can be divided out without losing a solution. Second — the step usually skipped — dividing by a cosine is legal only once you have checked that the cosine is not zero at any solution: say why the equation cannot hold where that cosine vanishes, and only then divide.

Then answer in the form asked for: the exact value first, written with arctan, and the three-decimal value after it. "Smallest positive" is doing work in that sentence — the equation has infinitely many solutions, and the principal value of arctan is what picks out the one wanted.

Getting the most out of it

Find the domain before you differentiate

Every logarithm, every even root and every arcsine restricts x before any rule is applied, and in this set that restriction usually carries its own mark. Write the domain down first, in standard interval notation, then check at the end that your derivative is not quietly claiming to exist somewhere the function does not.

Factor rather than expand

Almost every derivative here contains a common exponential, trigonometric or rational factor. Leaving it factored costs nothing now and pays for itself immediately: the points where a derivative vanishes are the zeros of its factors, and the sign of a derivative is far easier to read off a product than off an expanded sum.

Check it a second way

Exponentials are positive everywhere, so a sign you cannot justify is a sign that is wrong. A derivative supposed to be valid on (0, ∞) should not look comfortable at x = −1. A simplified trigonometric expression can be tested at one convenient angle against the unsimplified one. Each check takes seconds, and each catches the dropped minus sign that is otherwise invisible.

Answer the whole question

Domain restrictions, interval notation, the exact value as well as the decimal, the sentence saying which rule you used and why — these are separate marks from the derivative itself, and the cheapest ones on the page. They are also the ones most often left blank.

Want the solutions, or something more challenging?

The worksheet, the questions above and every explanation on this page stay free permanently. Three more PDFs exist for this topic — the worked answer key, a harder problem set, and the answer key to that. They come with the CEGEP Calculus I Solutions Bundle, which is what keeps the rest of the series free.

What else exists for Derivatives of Transcendental Functions

Three PDFs · 8 pages · all three are in the bundle below.

Answer key — 2 pages. All 6 questions worked step by step, including the restrictions and the justifications. Not a list of final answers.
Challenge problems — 4 pages, 4 problems. A separate sheet at exam-plus difficulty covering the same 5 concepts. Harder than anything on the free sheet.
Challenge answer key — 2 pages. Every challenge problem worked to the same standard, with the checks shown.
PDF, letter size, print-ready.

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Common questions

Is this worksheet really free?

Yes — the questions are on this page to read, and the PDF downloads directly, no email and no account. The one paid item is optional: the complete CEGEP Calculus I Solutions Bundle, which covers every set at this level.

Is this for 201-NYA-05 or 201-SN2-RE?

Both — they are the same course under two numbering systems. 201-NYA-05 is the legacy code and 201-SN2-RE the current one, and your college may use either. The content is set by competency 0M02, formerly 00UN, of the Sciences de la nature programme, which is what this worksheet is written against.

Why is ln(x²) equal to 2 ln|x| rather than 2 ln x?

Because the two sides would otherwise have different domains. ln(x²) is defined for every x except 0, since squaring makes the argument positive from either side, while 2 ln x is defined only for x greater than 0. The absolute value is what makes the identity true everywhere it needs to be. If a question has already restricted you to positive x you may drop it — but drop it because you checked, not out of habit.

What are the domains and ranges of arcsin and arctan?

arcsin is defined on [-1, 1] and returns an angle in [-π/2, π/2]. arctan is defined for every real number and returns an angle in (-π/2, π/2), open at both ends because the tangent has vertical asymptotes there. The devis names these two inverses among the concepts to be acquired in this course, which is a fair description of where the difficulty in this set sits.

Why does the derivative of arcsin exist on a smaller set than arcsin itself?

Because differentiating puts a square root in the denominator, and a denominator cannot be zero. arcsin needs its argument in the closed interval [-1, 1]; the derivative needs the strict inequality, so the endpoints drop out. Geometrically, those endpoints are where the graph has a vertical tangent, and a vertical tangent has no slope.

When should I use logarithmic differentiation?

In two situations. When the variable appears in both the base and the exponent, no ordinary rule applies — the power rule needs a fixed exponent and the exponential rule a fixed base — so taking logarithms is the only route. And when the expression is a long product, quotient and power stacked together, an ordinary derivative does exist, but the logarithm turns the whole thing into a sum and is far less work.

Do I need the chain rule for these questions?

Yes, constantly. The chain rule, the product rule, the quotient rule and implicit differentiation all come earlier in Calculus I and are assumed here rather than taught. What is new is the derivative of each transcendental function, and almost every question wraps one of those around an inner function whose derivative you have to remember to include.

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