This topic is about a single powerful tool — the derivative — and the three things the CSCA asks you to do with it: find the slope of a curve (and the tangent line at a point), decide where a function is rising or falling, and locate its peaks and valleys. All three are applications of one idea, so this first page builds that idea carefully and then delivers the most direct of the applications: the equation of a tangent line.
The derivative answers a question that ordinary algebra cannot: a straight line has one fixed slope everywhere, but a curve bends, so its steepness changes from point to point. How steep is a curve right here? The derivative is the machine that takes a function and returns its slope at every point. Before we can apply it, we need to know what it means and how to compute it for the polynomials the CSCA uses — so this page spends real time on the concept first, then the mechanics, then the payoff.
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What the derivative means
Start with what slope means for a straight line: rise over run, the change in y divided by the change in x. For a curve this fails, because the steepness is different at every point — near a valley the curve is nearly flat, on a steep climb it rises sharply. We want the slope at a single point, and here is the idea that makes that precise.
Pick the point on the curve where you want the slope, and a second point a little way along. The straight line through those two points — a secant — has an ordinary rise-over-run slope, and it roughly follows the curve's direction. Now slide the second point closer and closer to the first. The secant pivots, and its slope settles toward a single value: the slope of the tangent, the line that just grazes the curve at that one point. That limiting slope is the derivative at that point.
As the second point slides toward P, the secant's slope approaches the tangent's slope at P — that limiting value is the derivative.
So the derivative has two readings that are really the same thing. Geometrically, it is the slope of the curve (the slope of the tangent line) at a point. As a rate, it is the instantaneous rate of change — how fast the output is changing at that instant. We will lean on the geometric reading, because the three applications of this topic are all about the shape of a graph.
The derivative. The derivative of f, written f′(x), gives the slope of the curve y = f(x) at each point x — equivalently, the instantaneous rate of change of f. It is the limiting slope of secant lines as the second point approaches the first. Where f′(x) is large the curve is steep; where f′(x) = 0 the curve is momentarily flat.
The formal definition uses a limit — f′(x) is the limit of the secant slope [f(x + h) − f(x)] / h as h shrinks to 0. The CSCA does not ask you to compute derivatives from this limit; it expects you to use the shortcut rules below. But the limit is where those rules come from, and it is why "the second point slides in" is the right mental picture.
The power rule
Computing the derivative from the limit every time would be slow. Fortunately there is a shortcut for powers of x that does all the work, and since every polynomial is built from powers of x, it is essentially all you need for this topic. The rule: to differentiate xⁿ, bring the exponent down as a multiplier and reduce the exponent by one.
Power rule. If f(x) = xⁿ, then f′(x) = n·xⁿ⁻¹. Multiply by the old exponent, then subtract 1 from it. For example, the derivative of x³ is 3x², and the derivative of x² is 2x¹ = 2x.
We will take the power rule as a tool to memorize — its proof comes from the limit definition and is beyond what the CSCA asks — but it is easy to see it behaving sensibly. The derivative of x² is 2x, which grows as x grows: the parabola y = x² is gently sloped near the origin and steeper farther out, exactly as 2x describes.
Worked example · the power rule on single terms
Differentiate x⁵, and separately x³.
For x⁵, bring the 5 down and reduce the exponent: 5·x⁵⁻¹ = 5x⁴.
For x³, bring the 3 down and reduce: 3·x³⁻¹ = 3x².
So the derivatives are 5x⁴ and 3x². The pattern is mechanical: the old exponent becomes the coefficient, and the new exponent is one less.
Try it! · Check-in question
The derivative of x⁴ is ( ).
By the power rule, bring the 4 down and reduce the exponent: 4·x⁴⁻¹ = 4x³. The answer is C. Option A lowers the exponent but forgets to bring the 4 down in front. Option B raises the exponent (4 + 1 = 5) instead of lowering it. Option D brings the 4 down but then drops the power to 1, not to 4 − 1 = 3. The rule is: multiply by 4, then subtract 1 from the power, giving 4x³.
Try it! · Check-in question
Which function has the derivative 7x⁶? ( )
Differentiate each option and see which one gives 7x⁶. For x⁷, bring the 7 down and lower the power: 7x⁶. The answer is A. Option B gives 7·7x⁶ = 49x⁶: the 7 in front is not needed, because the power rule already brings a 7 down. Option C is the derivative of 7x⁶ (7·6x⁵ = 42x⁵), so it goes forward instead of backwards. Option D gives 6x⁵; the power must be one more than 6, so it must be 7.
Two special cases fall straight out of the rule. A plain x is x¹, so its derivative is 1·x⁰ = 1 — the line y = x has slope 1 everywhere, as expected. And a constant (like 7) has derivative 0: a horizontal line has zero slope, and nothing about a constant changes, so its rate of change is 0.
Try it! · Check-in question
The derivative of f(x) = π² is ( ).
π² is a fixed number (about 9.87). There is no x in it, so f is a constant function: its graph is a horizontal line, and its derivative is 0. The answer is B. Option A uses the power rule on π, as if π were the variable. The variable is x, and π never changes. Option C leaves the function as it is, without differentiating. Option D confuses a constant with x: the derivative of x is 1, but the derivative of a constant is 0.
Differentiating a whole polynomial
Real functions are sums of several terms with coefficients, like 2x³ + 3x² − 5. Two simple rules let the power rule handle these. First, a constant multiple stays out front: the derivative of c·f(x) is c·f′(x), so you differentiate the power and keep the coefficient. Second, a sum differentiates term by term: the derivative of a sum is the sum of the derivatives. Together they mean you can sweep across a polynomial, differentiating each term on its own.
Constant-multiple and sum rules. The derivative of c·xⁿ is c·n·xⁿ⁻¹ (keep the coefficient, apply the power rule). The derivative of a sum is the sum of the derivatives. So to differentiate a polynomial, differentiate each term separately and add the results.
Worked example · differentiating a polynomial
Differentiate f(x) = 2x³ + 3x² − 5.
Take the first term, 2x³: keep the 2, apply the power rule to x³ (which gives 3x²), so 2·3x² = 6x².
Take the second term, 3x²: keep the 3, differentiate x² to 2x, so 3·2x = 6x.
Take the constant −5: the derivative of any constant is 0.
Add the pieces: f′(x) = 6x² + 6x + 0 = 6x² + 6x.
So f′(x) = 6x² + 6x. Each term was handled on its own — coefficient kept, power rule applied, constant vanishing — then summed.
Brackets: multiply out first. Sometimes a polynomial is written as a product, such as x(x + 2). Multiply it out first: x(x + 2) = x² + 2x. Now every term is a power of x, so the rules above work: the derivative is 2x + 2. Do not differentiate each bracket and multiply the results. Here that would give 1 · 1 = 1, which is wrong. The derivative of a product is not the product of the derivatives.
Try it! · Check-in question
The derivative of f(x) = x³ − 4x + 1 is ( ).
Differentiate term by term: x³ → 3x², −4x → −4 (since the derivative of x is 1), and the constant 1 → 0. So f′(x) = 3x² − 4. The answer is D. Option A keeps an x on the −4x term; the derivative of x is 1, so −4x becomes −4. Option B brings the 3 down but does not lower the power on x³. Option C lowers the power but forgets to bring the 3 down in front. The constant drops out, leaving 3x² − 4.
Try it! · Check-in question
The derivative of f(x) = (x + 1)(x − 3) is ( ).
Multiply out first: (x + 1)(x − 3) = x² − 3x + x − 3 = x² − 2x − 3. Then differentiate term by term: 2x − 2. The answer is A. Option B differentiates each bracket (each gives 1) and multiplies: 1 · 1 = 1. That is not how a product works. Option C comes from an expanding slip: −3x − x = −4x instead of −3x + x = −2x. Option D expands correctly but then stops: x² − 2x − 3 is f(x) itself, not its derivative.
The slope at a particular point
The derivative f′(x) is itself a function — a formula that gives the slope at any x. To get the slope at one specific point, simply substitute that x-value into f′. The result is a single number: the steepness of the curve right there. This is the step that turns the general derivative into a concrete slope, and it is the heart of the tangent-line application coming next.
Slope at a point. The slope of y = f(x) at x = a is f′(a) — differentiate to get f′(x), then substitute x = a. A positive value means the curve rises there; negative means it falls; zero means it is momentarily flat (a peak, valley, or level spot).
Worked example · evaluating the slope
Find the slope of f(x) = x³ − 2x at the point where x = 2.
So the slope at x = 2 is 10 — the curve is rising steeply there. Notice the order: differentiate to get the slope formula, then plug in the point.
Differentiate first, then substitute. A common slip is to plug the x-value into the original f(x) (which gives the height, not the slope) or to substitute before differentiating. The slope at x = a is f′(a): find the derivative function first, then put x = a into it. Putting the number in too early leaves nothing to differentiate.
Try it! · Check-in question
For f(x) = x³ + x, the slope at x = −1 is ( ).
Differentiate: f′(x) = 3x² + 1. Substitute x = −1: 3·(−1)² + 1 = 3·1 + 1 = 4. The answer is C. Option A is f(−1) = −1 − 1 = −2, the height, not the slope. You also get −2 if you square −1 and get −1 (but (−1)² = +1). Option B forgets the +1: the derivative of x is 1, not 0. Option D writes f′(x) = x² + 1, forgetting to bring the 3 down.
Try it! · Check-in question
For f(x) = x² − 5x + 6, the slope at x = 0 is ( ).
Differentiate: f′(x) = 2x − 5. Substitute x = 0: 2·0 − 5 = −5. The answer is B. Option A is f(0) = 6, the height, not the slope. Option C puts x = 0 in too early: f(0) = 6 is a constant, and a constant differentiates to 0. Always differentiate first. Option D loses the minus sign: the derivative of −5x is −5, not 5.
The equation of a tangent line
Now the first big payoff. The tangent line at a point on a curve is the straight line that touches the curve there and matches its direction — sharing both the point and the slope. We have everything needed to find its equation: the slope is f′(a), and the line passes through the point (a, f(a)) on the curve. Feed those into the point-slope form of a line and you are done.
The tangent line shares the curve's point and slope at (a, f(a)): slope f′(a), passing through that point.
Equation of the tangent line at x = a. (1) Find the point: compute f(a) — the tangent passes through (a, f(a)). (2) Find the slope: compute f′(a). (3) Write the line in point-slope form y − f(a) = f′(a)·(x − a), then simplify to y = mx + b if asked.
Worked example · finding a tangent line
Find the equation of the tangent line to f(x) = x² at the point where x = 3.
Find the point: f(3) = 3² = 9, so the tangent passes through (3, 9).
Find the slope: f′(x) = 2x, so f′(3) = 2·3 = 6.
Write point-slope form: y − 9 = 6(x − 3).
Simplify: y − 9 = 6x − 18, so y = 6x − 9.
So the tangent line is y = 6x − 9. The two ingredients — the point (3, 9) and the slope 6 — are exactly what point-slope form needs.
Try it! · Check-in question
The tangent line to f(x) = x² + 1 at x = −1 is ( ).
Point: f(−1) = (−1)² + 1 = 2, so (−1, 2). Slope: f′(x) = 2x, so f′(−1) = −2. Point-slope: y − 2 = −2(x − (−1)) = −2(x + 1), so y = −2x − 2 + 2 = −2x. (Check: at x = −1, y = 2.) The answer is A. Option B writes (x − 1) instead of (x + 1). With a = −1, x − a is x + 1. Option C loses the minus sign of the slope and uses 2 instead of −2. Option D takes the point's height, 2, as the y-intercept (y = −2x + 2). But the line meets the curve at x = −1, not at x = 0.
Try it! · Check-in question
The tangent line to f(x) = x³ at x = 1 is ( ).
Point: f(1) = 1³ = 1, so (1, 1). Slope: f′(x) = 3x², so f′(1) = 3. Point-slope: y − 1 = 3(x − 1), so y = 3x − 3 + 1 = 3x − 2. The answer is D. Option A takes the point's height, 1, as the y-intercept. Option B writes f′(x) = x² (the 3 was not brought down), so it uses slope 1: y − 1 = x − 1. Option C moves the 1 to the wrong side: y = 3x − 3 − 1. Adding 1 to both sides gives 3x − 3 + 1.
Working backwards: from a slope to a point
Some questions turn the tangent problem around. They give you the slope and ask where on the curve the tangent has that slope. Use the same idea in reverse. The slope at x is f′(x), so set f′(x) equal to the given slope and solve for x. Then put each x into f(x) to get the point.
Two facts often give you the slope. Parallel lines have the same slope, so a tangent parallel to y = 3x + 1 has slope 3. A horizontal line has slope 0, so a horizontal tangent means f′(x) = 0.
Worked example · a point with a given slope
The tangent line to y = x³ at the point P has slope 12. Find P.
The slope at x is y′ = 3x². Set it equal to 12: 3x² = 12, so x² = 4.
Solve: x = 2 or x = −2. Do not forget the negative answer.
Find the y-values on the curve, using y = x³ (not y′): x = 2 gives y = 8, and x = −2 gives y = −8.
So P is (2, 8) or (−2, −8). The curve y = x³ has the same steepness at both points, so both answers are correct.
Try it! · Check-in question
At the point P on the curve y = x² − 3x, the tangent line is parallel to the line y = x + 5. The point P is ( ).
Parallel lines have the same slope, and y = x + 5 has slope 1. So set y′ = 2x − 3 equal to 1: 2x = 4, x = 2. Then y = 2² − 3·2 = −2, so P = (2, −2). The answer is C. Option A uses 5 as the slope: 2x − 3 = 5 gives x = 4 and y = 4. But 5 is the line's y-intercept, not its slope. Option B finds x = 2 but then gives the slope y′(2) = 1 as the y-value. The point must be on the curve, so use y = x² − 3x. Option D uses slope −1, which is the slope of a perpendicular line: 2x − 3 = −1 gives x = 1 and y = −2.
Your turn! · Practice questions
Practice · Question 1
The derivative of f(x) = (1/3)x³ − 2x² + 4 is ( ).
Term by term: (1/3)x³ → (1/3)·3x² = x², then −2x² → −2·2x = −4x, and the constant 4 → 0. So f′(x) = x² − 4x. The answer is B. Option A multiplies by 3 but does not lower the power on x³. Option C lowers the power but forgets to multiply by the 3 that comes down, so the 1/3 stays. Option D keeps the constant 4, but a constant differentiates to 0.
Practice · Question 2
An object moves along a line. Its position after t seconds is s(t) = t² + 2t metres. Its instantaneous rate of change at t = 3 is ( ) metres per second.
The instantaneous rate of change is the derivative. s′(t) = 2t + 2, so s′(3) = 6 + 2 = 8 metres per second. The answer is D. Option A is s(3) = 9 + 6 = 15, the position, not how fast it changes. Option B is the average rate from t = 0 to t = 3: (15 − 0) ÷ 3 = 5. That is the slope of a secant, not of the tangent. Option C forgets the +2: the derivative of 2t is 2, not 0.
Practice · Question 3
The tangent line to f(x) = x³ − x² at x = 2 is ( ).
Point: f(2) = 8 − 4 = 4, so (2, 4). Slope: f′(x) = 3x² − 2x, so f′(2) = 12 − 4 = 8. Point-slope: y − 4 = 8(x − 2), so y = 8x − 16 + 4 = 8x − 12. (Check: at x = 2, y = 4.) The answer is A. Option B takes the point's height, 4, as the y-intercept. Option C differentiates −x² as −x (the 2 was not brought down). That gives slope 12 − 2 = 10 and the line y = 10x − 16. Option D moves the 4 to the wrong side: y = 8x − 16 − 4.
Practice · Question 4
The tangent line to the curve y = x² + ax + 1 at x = 2 is horizontal. Then a = ( ).
A horizontal tangent has slope 0. The slope is y′ = 2x + a, so at x = 2: 4 + a = 0, which gives a = −4. The answer is C. Option A moves the 4 across without changing its sign. Option B writes y′ = x + a (the 2 was not brought down from x²), so 2 + a = 0. Option D sets the height y(2) = 4 + 2a + 1 equal to 0. "Horizontal" is about the slope, not the height.
Things to remember
Concept
What to know
The derivative
f′(x) is the slope of the curve at each point (the instantaneous rate of change), the limiting slope of secants becoming a tangent.
Power rule
xⁿ differentiates to n·xⁿ⁻¹: bring the exponent down, reduce it by one.
Constant & sum rules
Keep coefficients (c·xⁿ → c·n·xⁿ⁻¹); differentiate sums term by term. A constant differentiates to 0; x differentiates to 1.
Slope at a point
The slope at x = a is f′(a): differentiate first, then substitute. Positive rises, negative falls, zero is flat.
Tangent line
At x = a: point (a, f(a)), slope f′(a), then y − f(a) = f′(a)(x − a). Simplify to y = mx + b.
Slope given, point unknown
Solve f′(x) = the slope, then find y from f(x). Parallel lines have equal slopes; a horizontal tangent has slope 0. Look for more than one answer.
Brackets
Multiply out a product before you differentiate.
Order matters
Always differentiate before substituting a value — plugging in first leaves the height, not the slope.
