Difficulty: Intermediate | Prerequisites: Chain Rule notes, basic algebra, trigonometry
This set of notes pulls together every differentiation technique covered so far in the course: power rule, product rule, quotient rule, chain rule, and trig derivatives. The problems in this section require you to recognise which rule (or combination of rules) applies and execute it cleanly. Most exam questions at this level mix two or more rules in a single problem, so fluency with each rule individually is the prerequisite, and combining them is the skill being tested.
Differentiation has about half a dozen core rules, and nearly every problem you will see asks you to combine two or more of them in a single question. The skill is not memorising the rules (you probably already have) but recognising which rule applies to which piece of the function, then executing cleanly without sign errors or missing terms.
Power rule
If f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. In simple terms, bring the exponent down as a coefficient and reduce the exponent by one. Works for any real exponent, including negatives and fractions.
Constant multiple rule
d/dx[c · f(x)] = c · f'(x). A constant factor passes straight through differentiation. Think of it as: constants are boring, so they just come along for the ride.
Sum and difference rule
d/dx[f(x) ± g(x)] = f'(x) ± g'(x). You differentiate each term separately. In simple terms, the derivative of a sum is the sum of the derivatives.
Product rule
If h(x) = f(x) · g(x), then h'(x) = f'(x) · g(x) + f(x) · g'(x). Think of it as: "first times derivative of the second, plus second times derivative of the first." Some students remember it as "derivative of the first times the second, plus the first times the derivative of the second."
Quotient rule
If h(x) = f(x)/g(x), then h'(x) = [f'(x)·g(x) − f(x)·g'(x)] / [g(x)]². Think of it as: "lo d-hi minus hi d-lo, over lo-lo" (low times derivative of high, minus high times derivative of low, all over low squared). The minus sign is the main source of errors.
Horizontal tangent line
A point where f'(x) = 0. The tangent line is flat (slope zero) at that point. Finding horizontal tangent lines means solving f'(x) = 0 for x.
Tangent to a curve
The line that just touches a curve at a point, with the same slope as the curve at that point. Its equation is y − f(a) = f'(a)(x − a).
Many functions need algebraic rewriting before any rule applies cleanly. This is one of the most important habits to build.
Negative exponents: 7/x³ becomes 7x⁻³. Then the power rule gives −21x⁻⁴, which you rewrite as −21/x⁴.
Fractional exponents: √x = x^(1/2), so d/dx[5√x] = 5 · (1/2)x⁻^(1/2) = 5/(2√x).
Constants: e² is just a number (roughly 7.389). Its derivative is zero. Students sometimes try to apply the exponential rule to it.
Example: f(x) = 3x² − 5√x + 7/x³ − e². Rewrite as 3x² − 5x^(1/2) + 7x⁻³ − e². f'(x) = 6x − (5/2)x⁻^(1/2) − 21x⁻⁴ = 6x − 5/(2√x) − 21/x⁴.
Use the product rule when two variable expressions are multiplied together.
Example: y = x² sin(x). Let f = x², g = sin(x). y' = 2x sin(x) + x² cos(x).
Note that if either factor also requires the chain rule (say sin(5x + 3) instead of sin(x)), you apply the chain rule to that factor's derivative as part of the product rule.
Use the quotient rule when one variable expression is divided by another. The formula is h' = (f'g − fg')/g².
Example: g(x) = tan(x)/(1 − 2x). Numerator derivative: sec²(x). Denominator derivative: −2. g'(x) = [sec²(x)(1 − 2x) − tan(x)(−2)] / (1 − 2x)² g'(x) = [(1 − 2x)sec²(x) + 2 tan(x)] / (1 − 2x)².
Alternative approach: if the denominator is simple, you can sometimes rewrite as a product and use the product rule instead. For g(x) = 5/(3x²) = (5/3)x⁻², the power rule alone gives g'(x) = (5/3)(−2)x⁻³ = −10/(3x³).
Most problems at this level layer two or more rules.
Example: h(x) = 4 cos³(5x). Three layers here: the constant 4, the cube, the cosine of 5x. h'(x) = 4 · 3 cos²(5x) · (−sin(5x)) · 5 = −60 cos²(5x) sin(5x).
Example: y = (x⁴ − 7x)¹⁵. y' = 15(x⁴ − 7x)¹⁴ · (4x³ − 7).
Set f'(x) = 0 and solve. For y = √3 · x − 2 sin(x) on [0, 2π): y' = √3 − 2 cos(x). Set √3 − 2 cos(x) = 0, so cos(x) = √3/2. x = π/6 and x = 11π/6 on [0, 2π).
For f(x) = cos(2x) at x = π/6: f(π/6) = cos(π/3) = 1/2. f'(x) = −2 sin(2x), so f'(π/6) = −2 sin(π/3) = −2(√3/2) = −√3. Tangent line: y − 1/2 = −√3(x − π/6).
For f(x) = x³ − 3x² + x + 4, find where the slope is 10. f'(x) = 3x² − 6x + 1. Set 3x² − 6x + 1 = 10, so 3x² − 6x − 9 = 0, giving x² − 2x − 3 = 0, so (x − 3)(x + 1) = 0. x = 3 or x = −1. The points are (3, 7) and (−1, 5).
For f(x) = k/x tangent to y = −(3/4)x + 3: f'(x) = −k/x². At the point of tangency, the slope equals −3/4, so −k/x² = −3/4, giving k = 3x²/4. Also, k/x = −(3/4)x + 3. Substituting k = 3x²/4: (3x/4) = −(3/4)x + 3, so (3x/2) = 3, x = 2, k = 3.
These give you a table of values for f, g, f', g' at specific x values and ask you to evaluate a derivative of a combination at a point. No formula is given. You just plug into the rule.
Example (product rule): k(x) = f(x)·g(x). k'(4) = f'(4)·g(4) + f(4)·g'(4) = (−2)(2) + (3)(3) = −4 + 9 = 5.
Example (quotient rule): m(x) = f(x)/g(x). m'(4) = [f'(4)·g(4) − f(4)·g'(4)] / [g(4)]² = [(−2)(2) − (3)(3)] / 4 = (−4 − 9)/4 = −13/4.
Example (chain rule): p(x) = f(g(x)). p'(2) = f'(g(2)) · g'(2) = f'(4) · 9 = (−2)(9) = −18.
\frac{d}{dx}[x^n] = nx^{n-1}\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)\frac{d}{dx}[f \pm g] = f' \pm g'(fg)' = f'g + fg'\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}\frac{d}{dx}[\sin x] = \cos x \qquad \frac{d}{dx}[\cos x] = -\sin x\frac{d}{dx}[\tan x] = \sec^2 x \qquad \frac{d}{dx}[\sec x] = \sec x \tan xf''(x) = \frac{d}{dx}[f'(x)]Every rate-of-change problem in science and engineering comes back to these rules. Velocity is the derivative of position. Acceleration is the second derivative. In economics, marginal cost and marginal revenue are derivatives of cost and revenue functions. The quotient rule appears any time you differentiate a ratio (efficiency, density, per-capita quantities). These rules are not abstract exercises; they are the machinery behind every optimisation and modelling problem you will encounter.
Students mix up the sign in the quotient rule. The formula is f'g minus fg', not the other way round. Swapping them flips the sign of the entire answer.
Students treat e² as an exponential function and try to differentiate it as eˣ. It is a constant (approximately 7.389), so its derivative is zero.
When rewriting radicals, students forget that √x = x^(1/2), not x². This leads to the exponent going the wrong direction.
After applying the quotient rule, students sometimes leave negative exponents or complex fractions in the final answer. The worksheet instructions explicitly say to clear these out.
In table-based problems, students evaluate f'(x) · g'(x) instead of f'(g(x)) · g'(x) for the chain rule composition p(x) = f(g(x)). The inner function's value, not its derivative, feeds into f'.
⚠️ Exams at this level test rule selection as much as execution. A problem that looks like it needs the quotient rule might simplify to a power rule problem if you rewrite first.
⚠️ "Find the second derivative" questions are common and carry higher marks. They almost always require the product rule on the first derivative.
⚠️ "Find where the tangent line is horizontal" is a standard question. It tests both differentiation and algebraic solving.
⚠️ Table-based derivative problems test conceptual understanding. You cannot compute these mechanically; you must know what the product rule, quotient rule, and chain rule mean.
⚠️ "Find k such that..." problems test whether you can set up equations from a tangency condition (equal y-values and equal slopes at the same x).
True or False: The derivative of 7/x³ is −7/x². (False: rewrite as 7x⁻³, so the derivative is −21x⁻⁴ = −21/x⁴.)
Fill in the blank: d/dx[x² sin(x)] = 2x sin(x) + ___. (x² cos(x))
True or False: e² is a constant, so its derivative is 0. (True.)
Fill in the blank: The quotient rule formula is (f'g − fg') / ___. (g²)
True or False: To find f''(x) for f(x) = 2x³ − 5x² + 9, you first find f'(x) = 6x² − 10x, then differentiate again to get f''(x) = 12x − 10. (True.)
Q: Find f'(x) for f(x) = 3x² − 5√x + 7/x³ − e².
A: Rewrite as 3x² − 5x^(1/2) + 7x⁻³ − e². Then f'(x) = 6x − (5/2)x⁻^(1/2) − 21x⁻⁴ = 6x − 5/(2√x) − 21/x⁴.
Q: Find dy/dx for y = (x⁴ − 7x)¹⁵.
A: y' = 15(x⁴ − 7x)¹⁴ · (4x³ − 7)
Q: Find h'(x) for h(x) = 4 cos³(5x).
A: h'(x) = 4 · 3 cos²(5x) · (−sin(5x)) · 5 = −60 cos²(5x) sin(5x)
Q: Find dy/dx for y = x² sin(x).
A: Product rule. y' = 2x sin(x) + x² cos(x)
Q: Find f'(x) for f(x) = sin(5x + 3).
A: f'(x) = cos(5x + 3) · 5 = 5 cos(5x + 3)
Q: Find g'(x) for g(x) = 5/(3x²).
A: Rewrite as (5/3)x⁻². Then g'(x) = (5/3)(−2)x⁻³ = −10/(3x³)
Q: Find g'(x) for g(x) = tan(x)/(1 − 2x).
A: Quotient rule. g'(x) = [(1 − 2x) sec²(x) + 2 tan(x)] / (1 − 2x)²
Q: Find dy/dx for y = √(3x) + 3/x².
A: Rewrite as (3x)^(1/2) + 3x⁻². Then y' = (1/2)(3x)⁻^(1/2) · 3 − 6x⁻³ = 3/(2√(3x)) − 6/x³. Simplify the first term: √3/(2√x) − 6/x³.
Q: Find f''(x) for f(x) = 2x³ − 5x² + 9.
A: f'(x) = 6x² − 10x. f''(x) = 12x − 10.
Q: Find g''(x) for g(x) = cos(2x³).
A: g'(x) = −sin(2x³) · 6x² = −6x² sin(2x³). Then g''(x) requires the product rule on −6x² and sin(2x³): g''(x) = −12x sin(2x³) − 6x² · cos(2x³) · 6x² = −12x sin(2x³) − 36x⁴ cos(2x³).
Q: Find the points where y = √3 · x − 2 sin(x) has a horizontal tangent on [0, 2π).
A: y' = √3 − 2 cos(x) = 0. cos(x) = √3/2. x = π/6 and x = 11π/6. The points are (π/6, √3 · π/6 − 1) and (11π/6, 11√3π/6 + 1).
Q: Write the tangent line to f(x) = cos(2x) at x = π/6.
A: f(π/6) = cos(π/3) = 1/2. f'(x) = −2 sin(2x), f'(π/6) = −2 sin(π/3) = −√3. Tangent line: y − 1/2 = −√3(x − π/6).
Q: Given f(x) = x³ − 3x² + x + 4, find points where the tangent has slope 10.
A: f'(x) = 3x² − 6x + 1 = 10. Then 3x² − 6x − 9 = 0, so x² − 2x − 3 = 0, (x − 3)(x + 1) = 0. x = 3: f(3) = 27 − 27 + 3 + 4 = 7. x = −1: f(−1) = −1 − 3 − 1 + 4 = −1. Points: (3, 7) and (−1, −1).
Q: Using the table (f(2) = 5, g(2) = 4, f'(2) = −3, g'(2) = 9, f(4) = 3, g(4) = 2, f'(4) = −2, g'(4) = 3), find h'(2) for h(x) = 3f(x) − 2g(x).
A: h'(x) = 3f'(x) − 2g'(x). h'(2) = 3(−3) − 2(9) = −9 − 18 = −27.
Q: Using the same table, find k'(4) for k(x) = f(x)·g(x).
A: Product rule. k'(4) = f'(4)·g(4) + f(4)·g'(4) = (−2)(2) + (3)(3) = −4 + 9 = 5.
Q: Using the same table, find p'(2) for p(x) = f(g(x)).
A: Chain rule. p'(2) = f'(g(2)) · g'(2) = f'(4) · 9 = (−2)(9) = −18.
These differentiation rules are the foundation for optimisation problems (finding maxima and minima), curve sketching (using the first and second derivatives), and related rates. The chain rule in particular reappears as u-substitution in integration. The product and quotient rules become essential when working with logarithmic differentiation and implicit differentiation in the next chapter.
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