Difficulty: Intermediate | Prerequisites: Limits and continuity
Derivatives measure instantaneous rate of change, which is the slope of the curve at a single point. This section covers the mechanical rules for computing them (power rule, product rule, quotient rule, chain rule) and the technique of implicit differentiation for equations that cannot be solved for y. Mastering these rules is mostly pattern recognition: once you see which rule applies, the algebra follows.
Derivative, f'(x)
The instantaneous rate of change of f at x. Geometrically, it is the slope of the tangent line to the curve at that point. In simple terms, it tells you how fast the function is changing right now.
Tangent line
The line that just touches the curve at a point and has the same slope as the curve there. Its equation is y − f(a) = f'(a)(x − a).
Normal line
The line perpendicular to the tangent line at the point of tangency. Its slope is −1/f'(a), the negative reciprocal of the tangent slope.
Product rule
For y = f(x) · g(x): y' = f'(x) · g(x) + f(x) · g'(x). Think of it as: "derivative of the first times the second, plus the first times the derivative of the second."
Quotient rule
For y = f(x)/g(x): y' = (f'(x) · g(x) − f(x) · g'(x)) / [g(x)]². Think of it as: "low d-high minus high d-low, over the square of what's below."
Chain rule
For y = f(g(x)): y' = f'(g(x)) · g'(x). Differentiate the outer function, keep the inner function unchanged, then multiply by the derivative of the inner function.
Implicit differentiation
A technique for finding dy/dx when y is not isolated on one side of the equation. Differentiate both sides with respect to x, treat y as a function of x (so every time you differentiate a y-term, attach a dy/dx factor), then solve for dy/dx.
If f(x) = xⁿ, then f'(x) = nxⁿ⁻¹.
The derivative of a constant is 0.
The derivative of eˣ is eˣ. The derivative of ln(x) is 1/x.
The derivative of sin(x) is cos(x). The derivative of cos(x) is −sin(x). The derivative of tan(x) is sec²(x).
Given f(x) = 3x² − 2x at x = 2:
f'(x) = 6x − 2, so f'(2) = 10. The slope of the curve at x = 2 is 10.
f(2) = 3(4) − 4 = 8. The point is (2, 8).
Tangent line: y − 8 = 10(x − 2).
Normal line slope = −1/10. Normal line: y − 8 = −(1/10)(x − 2).
For f(x) = x/(x + 1) at x = 2:
Use the quotient rule: f'(x) = (1·(x + 1) − x·1)/(x + 1)² = 1/(x + 1)².
f'(2) = 1/9.
To find where the tangent to y = 4x² − 3x is parallel to y = 2x − 5, set y' = 2. Since y' = 8x − 3, solve 8x − 3 = 2 to get x = 5/8. The point is (5/8, −5/16), and the tangent line is y + 5/16 = 2(x − 5/8).
The formula: if y = f(x)/g(x), then y' = [f'(x)·g(x) − f(x)·g'(x)] / [g(x)]².
y = (2x² − x)/(x² − x). Find y'(3).
y' = [(4x − 1)(x² − x) − (2x² − x)(2x − 1)] / (x² − x)².
At x = 3: y'(3) = [(11)(6) − (15)(5)] / 6² = (66 − 75)/36 = −9/36 = −1/4.
f(x) = (x + 3)/(x − 2). Find the normal line at (1, −4).
f'(x) = [1·(x − 2) − (x + 3)·1] / (x − 2)² = −5/(x − 2)².
f'(1) = −5/1 = −5. The tangent slope is −5, so the normal slope is 1/5.
Normal line: y + 4 = (1/5)(x − 1).
y = e⁻²ˣ/(x + 3). This combines the quotient rule with the chain rule.
y' = [−2e⁻²ˣ(x + 3) − e⁻²ˣ·1] / (x + 3)² = [−2e⁻²ˣ(x + 3) − e⁻²ˣ] / (x + 3)².
y = 3x/(x² − 2). Find the slope of the tangent at (2, 3).
y' = [3(x² − 2) − (2x)(3x)] / (x² − 2)² = [3x² − 6 − 6x²] / (x² − 2)² = (−3x² − 6)/(x² − 2)².
y'(2) = (−12 − 6)/4 = −18/4 = −9/2.
The chain rule applies whenever you differentiate a composite function, meaning a function inside another function. Differentiate the outer layer, leave the inner layer alone, then multiply by the derivative of the inner layer.
f(x) = sin⁴(5x). Rewrite as [sin(5x)]⁴.
Outer: power rule gives 4[sin(5x)]³.
Inner (sin): derivative is cos(5x).
Innermost (5x): derivative is 5.
Result: f'(x) = 20 sin³(5x) · cos(5x).
f(x) = tan(3x). f'(x) = sec²(3x) · 3 = 3 sec²(3x).
f(x) = cos⁻¹(7x), i.e. arccos(7x). f'(x) = −1/√(1 − 49x²) · 7 = −7/√(1 − 49x²).
f(x) = arcsin(2x). f'(x) = 2/√(1 − 4x²).
y = √(x² − 5) = (x² − 5)^(1/2).
Outer: (1/2)(x² − 5)^(−1/2).
Inner: 2x.
Result: y' = x/√(x² − 5).
y = x³ · ln(x − 8). Use the product rule first, then chain rule on ln(x − 8).
y' = 3x² · ln(x − 8) + x³ · 1/(x − 8) = 3x² ln(x − 8) + x³/(x − 8).
y = log₅(3x² + 4x). Use the log derivative rule: d/dx[logₐ(u)] = u'/(u · ln a).
y' = (6x + 4)/[(3x² + 4x) · ln 5].
f(x) = (7x + √(x² + 3))⁶.
Outer: 6(7x + √(x² + 3))⁵.
Inner derivative: 7 + x/√(x² + 3).
Result: f'(x) = 6(7x + √(x² + 3))⁵ · (7 + x/√(x² + 3)).
f(x) = x cos(x). Product rule: f'(x) = cos(x) − x sin(x).
f(x) = (3x + 1)⁴(x − 1). Product rule with chain rule on the first factor.
f'(x) = 4(3x + 1)³ · 3 · (x − 1) + (3x + 1)⁴ · 1.
At x = −1: f'(−1) = 12(−2)³(−2) + (−2)⁴ = 192 + 16 = 208.
The tangent line at (−1, −32): y + 32 = 208(x + 1).
Use implicit differentiation when y is tangled with x in an equation you cannot (or do not want to) solve for y. Differentiate every term with respect to x. Whenever you differentiate a term containing y, multiply by dy/dx (because y is a function of x). Then isolate dy/dx.
5x + xy² = y⁴ + 7. Find dy/dx at (−1, 3).
Differentiate: 5 + y² + 2xy(dy/dx) = 4y³(dy/dx).
Collect dy/dx terms: 2xy(dy/dx) − 4y³(dy/dx) = −5 − y².
(dy/dx)(2xy − 4y³) = −y² − 5.
dy/dx = (−y² − 5)/(2xy − 4y³).
At (−1, 3): dy/dx = (−9 − 5)/(−6 − 108) = −14/(−114) = 7/57.
xy + y² + x³ = 7. Find dy/dx.
Differentiate: y + x(dy/dx) + 2y(dy/dx) + 3x² = 0.
(dy/dx)(x + 2y) = −3x² − y.
dy/dx = (−3x² − y)/(x + 2y).
3x² − xy − y³ = 5. Find dy/dx.
Differentiate: 6x − y − x(dy/dx) − 3y²(dy/dx) = 0.
(dy/dx)(−x − 3y²) = −6x + y.
dy/dx = (−6x + y)/(−x − 3y²) = (6x − y)/(x + 3y²).
2x²y + y² = 2x + 13. Find the tangent slope at (3, 1).
Differentiate: 4xy + 2x²(dy/dx) + 2y(dy/dx) = 2.
(dy/dx)(2x² + 2y) = 2 − 4xy.
dy/dx = (2 − 4xy)/(2x² + 2y).
At (3, 1): dy/dx = (2 − 12)/(18 + 2) = −10/20 = −1/2.
Students forget to multiply by dy/dx when differentiating y-terms during implicit differentiation. Every time you differentiate a y-term, the chain rule produces a dy/dx factor. Missing it is the single most common error.
When using the quotient rule, students frequently get the numerator subtraction backwards. It is f'g − fg', not fg' − f'g. The derivative of the top goes first.
Students apply the chain rule to simple polynomials where it is not needed, overcomplicating the work. If there is no function-inside-a-function structure, the power rule alone suffices.
Students confuse the tangent line with the normal line. The tangent line has slope f'(a). The normal line has slope −1/f'(a). Read the question carefully to see which one is asked for.
⚠️ The exam will almost certainly include a quotient rule problem and a chain rule problem. Expect composite trig functions like sin⁴(5x) or arcsin(2x).
⚠️ Implicit differentiation questions typically ask for dy/dx at a specific point. Do not forget to substitute the point only after you have the general dy/dx expression.
⚠️ "Find the equation of the tangent line" and "find the equation of the normal line" are high-frequency exam questions. You need the slope (from the derivative) and the point (from the original function).
⚠️ Problems that combine the product rule with the chain rule (e.g. f(x) = (3x + 1)⁴(x − 1)) are common. Take them one layer at a time.
⚠️ The instantaneous rate of change is just f'(a). If the question asks for it at a specific x, differentiate and substitute.
⚠️ Second derivatives (f''(x)) may appear. For the quotient rule applied twice (as in problem 53 of the review), keep your algebra tidy and do not skip steps.
True or false: The derivative of sin(3x) is cos(3x). (False. It is 3cos(3x), because of the chain rule.)
Fill in the blank: The normal line has slope ___ when the tangent slope is 4. (−1/4)
True or false: In implicit differentiation, when you differentiate y² with respect to x, you get 2y. (False. You get 2y · dy/dx.)
Fill in the blank: The quotient rule numerator is f'g ___ fg'. (minus)
True or false: The derivative of ln(x − 8) is 1/(x − 8). (True. The inner derivative is 1, so the chain rule does not add anything extra.)
Q: Find f'(x) for f(x) = sin⁴(5x).
A: Rewrite as [sin(5x)]⁴. Apply the chain rule in layers: 4[sin(5x)]³ · cos(5x) · 5 = 20 sin³(5x) cos(5x).
Q: Find dy/dx for 5x + xy² = y⁴ + 7 at (−1, 3).
A: Differentiate implicitly: 5 + y² + 2xy(dy/dx) = 4y³(dy/dx). Solve for dy/dx = (−y² − 5)/(2xy − 4y³). At (−1, 3): dy/dx = −14/(−114) = 7/57.
Q: Find the equation of the tangent line to f(x) = (3x + 1)⁴(x − 1) at x = −1.
A: f'(x) = 4(3x + 1)³(3)(x − 1) + (3x + 1)⁴. At x = −1: f'(−1) = 12(−2)³(−2) + (−2)⁴ = 192 + 16 = 208. f(−1) = (−2)⁴(−2) = −32. Tangent line: y + 32 = 208(x + 1).
Q: Find the instantaneous rate of change of f(x) = x²/(x + 1) at x = −3.
A: f'(x) = [2x(x + 1) − x²]/(x + 1)² = (x² + 2x)/(x + 1)². At x = −3: f'(−3) = (9 − 6)/4 = 3/4.
Q: Find the slope of the tangent line to 2x²y + y² = 2x + 13 at (3, 1).
A: Implicit differentiation gives dy/dx = (2 − 4xy)/(2x² + 2y). At (3, 1): dy/dx = (2 − 12)/(18 + 2) = −10/20 = −1/2.
Derivative rules are the mechanical tools you need for everything that follows. Motion problems (velocity and acceleration) rely on taking first and second derivatives of position functions. Curve sketching uses the first derivative to find intervals of increase/decrease and the second derivative for concavity. Optimisation problems set the derivative equal to zero to find critical points. All of these applications are covered in the companion study notes on Applications of Derivatives.
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