Difficulty: Introductory
Prerequisites: Basic algebra (exponents, polynomial expressions, function notation)
Derivatives are the central idea in differential calculus: they measure how a function's output changes as its input changes. This set of problems covers the foundational rules you will use to differentiate almost every function you meet in the course, from simple power functions to polynomials, roots, and reciprocals. If you can apply the power rule, the constant multiple rule, and the sum/difference rule fluently, you have the toolkit for nearly everything that follows in Brief Calculus.
The power rule lets you differentiate any term of the form xⁿ by bringing the exponent down and reducing it by one. Constants multiply through, sums and differences split apart, and standalone constants vanish. Rewriting roots and reciprocals as fractional or negative exponents turns them into straightforward power-rule problems.
Derivative
The rate of change of a function with respect to its variable. Notation: f'(x) or dy/dx.
In simple terms, this means: how fast the output is changing at any given input.
Power rule
If f(x) = xⁿ, then f'(x) = n·xⁿ⁻¹. Bring the exponent down as a coefficient, then subtract one from the exponent.
Think of it as: "drop, then reduce."
Constant multiple rule
The derivative of a constant times a function equals the constant times the derivative of the function: d/dx [c·f(x)] = c·f'(x).
In simple terms, this means: constants pass straight through the differentiation.
Sum and difference rule
The derivative of a sum (or difference) of functions equals the sum (or difference) of their individual derivatives: d/dx [f(x) ± g(x)] = f'(x) ± g'(x).
Think of it as: differentiate each term on its own, then recombine.
Derivative of a constant
The derivative of any constant is zero: d/dx (c) = 0.
In simple terms, this means: a flat horizontal line has no slope.
Fractional exponent
A way of writing roots as powers. For example, √x = x^(1/2) and ∛x = x^(1/3).
Think of it as: rewriting the root so you can use the power rule on it.
Negative exponent
A way of writing reciprocals as powers. For example, 1/x⁶ = x⁻⁶.
In simple terms, this means: moving a term from the denominator to the numerator flips the sign of its exponent.
If f(x) = xⁿ, then f'(x) = n·xⁿ⁻¹.
Multiply the function by the current exponent, then reduce the exponent by one.
Works for any real exponent: positive integers, fractions, negatives, decimals.
Worked examples from the problem set:
f(x) = x⁶ → n = 6 → f'(x) = 6x⁵
f(x) = x⁷ → n = 7 → f'(x) = 7x⁶
d/dx [c·f(x)] = c · d/dx [f(x)]
Pull the constant out, differentiate the rest, then multiply the constant back in.
Worked examples:
y = 10x → c = 10, f(x) = x → d/dx(10x) = 10 · d/dx(x) = 10 · 1 = 10
y = 7x → c = 7, f(x) = x → d/dx(7x) = 7 · 1 = 7
d/dx (c) = 0 for any constant c.
A constant has no variable part, so its rate of change is always zero.
Worked example:
y = 10 → dy/dx = 0
d/dx [f(x) + g(x)] = f'(x) + g'(x)
Differentiate each term independently, then add (or subtract) the results.
Signs carry through: a term subtracted in the original stays subtracted in the derivative.
Worked example (Problem 11):
y = 2x³ - 5x² + 8x + 9
Differentiate term by term:
First term: d/dx(2x³) = 2 · 3x² = 6x²
Second term: d/dx(5x²) = 5 · 2x = 10x
Third term: d/dx(8x) = 8
Fourth term: d/dx(9) = 0 (constant)
Combine with original signs: dy/dx = 6x² - 10x + 8
Note: the "+0" from the constant is dropped.
Worked example (Problem 12):
y = 2x³ - 13x² + 16x + 6
d/dx(2x³) = 6x²
d/dx(13x²) = 26x
d/dx(16x) = 16
d/dx(6) = 0
dy/dx = 6x² - 26x + 16 (note the signs: the second term was subtracted, so -26x; the third was added, so +16)
The power rule works identically with fractional and negative exponents. The key step is rewriting the function first.
Roots as fractional exponents:
√x = x^(1/2)
∛x = x^(1/3)
Then apply the power rule as normal. The result often looks cleaner rewritten back in root form.
Worked example (Problem 13):
y = 20√x = 20x^(1/2)
d/dx(20x^(1/2)) = 20 · (1/2) · x^(1/2 - 1) = 10x^(-1/2)
Rewrite: 10x^(-1/2) = 10/√x
Worked example (Problem 14):
y = 12√x = 12x^(1/2)
d/dx = 12 · (1/2) · x^(-1/2) = 6x^(-1/2) = 6√x / x = 6/√x
Worked example (Problem 15) with cube roots:
y = -4.2x^(1/3)
d/dx = -4.2 · (1/3) · x^(1/3 - 1) = -1.4x^(-2/3)
Calculation check: (-4.2)(1/3) = -1.4. The exponent becomes 1/3 - 1 = -2/3.
Negative exponents (reciprocals):
1/xⁿ = x⁻ⁿ
Rewrite the function with a negative exponent, then use the power rule.
Worked example (Problem 16):
y = -30/x⁶ = -30x⁻⁶
d/dx = -30 · (-6) · x^(-6-1) = 180x⁻⁷
Rewrite: 180x⁻⁷ = 180/x⁷
Note: the two negatives (the -30 coefficient and the -6 exponent) multiply to give a positive 180.
\frac{d}{dx}[x^n] = n \cdot x^{n-1}\frac{d}{dx}[c \cdot f(x)] = c \cdot \frac{d}{dx}[f(x)]\frac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)\frac{d}{dx}[c] = 0Conversion reminders:
√x = x^(1/2)
∛x = x^(1/3)
1/xⁿ = x⁻ⁿ
Derivatives measure rates of change, so these rules appear everywhere a quantity is changing. If a company's revenue is modelled as a polynomial in time, the derivative gives the rate at which revenue is growing or shrinking at any moment. In physics, if position is a function of time, the derivative is velocity. Engineers use derivatives to find slopes of curves when designing roads, roller coasters, and optical lenses.
Students often forget to subtract 1 from the exponent after applying the power rule. If f(x) = x⁶, the derivative is 6x⁵, not 6x⁶.
Students sometimes try to "differentiate" a constant and get 1 instead of 0. The derivative of any standalone number is always zero.
When differentiating a polynomial, students sometimes lose track of the signs. Each term's sign (+ or -) carries through unchanged. Write out each term's derivative separately before combining.
Students forget to rewrite roots and reciprocals as exponents before differentiating. You cannot apply the power rule to √x directly; rewrite it as x^(1/2) first.
⚠️ The power rule is the most heavily tested differentiation rule in introductory calculus. Expect it on every exam.
⚠️ Polynomial differentiation (sum/difference rule combined with constant multiple rule) is a near-certain exam question. Practise until the term-by-term process is automatic.
⚠️ Rewriting roots as fractional exponents is a common exam trap. If you see √x or ∛x, your first move is always to convert to x^(1/2) or x^(1/3).
⚠️ Negative exponent problems (like y = -30/x⁶) test whether you can rewrite before differentiating. Watch for sign errors when two negatives multiply to a positive.
True or false: The derivative of x⁴ is 4x⁴.
False. It is 4x³ (subtract 1 from the exponent).
Fill in the blank: d/dx [5x³] = ___
15x²
True or false: The derivative of the number 42 is 42.
False. The derivative of any constant is 0.
Fill in the blank: √x rewritten as a power is x^(___)
1/2
True or false: d/dx [f(x) + g(x)] = f'(x) · g'(x).
False. The sum rule gives f'(x) + g'(x), not a product.
Q: Differentiate f(x) = x⁹.
A: f'(x) = 9x⁸. Apply the power rule: bring down the 9, reduce the exponent to 8.
Q: Find dy/dx if y = 3x⁴ - 7x² + 2x - 11.
A: dy/dx = 12x³ - 14x + 2. Differentiate each term: 3·4x³ = 12x³, 7·2x = 14x (keep the minus), 2·1 = 2, and the constant -11 drops out.
Q: Differentiate y = 8√x.
A: Rewrite as y = 8x^(1/2). Then dy/dx = 8 · (1/2) · x^(-1/2) = 4x^(-1/2) = 4/√x.
Q: Find dy/dx if y = -5/x³.
A: Rewrite as y = -5x⁻³. Then dy/dx = -5 · (-3) · x⁻⁴ = 15x⁻⁴ = 15/x⁴.
Q: What is the derivative of y = 6x?
A: dy/dx = 6. The function is a constant times x (which has exponent 1). Applying the power rule to x gives 1, so 6 · 1 = 6.
These basic differentiation rules are the building blocks for everything else in calculus. The product rule and quotient rule (covered next) handle functions that are multiplied or divided, rather than simply added. The chain rule extends the power rule to composite functions like (3x + 1)⁵. Later, you will use derivatives to find maximum and minimum values of functions, which is the foundation of optimisation problems in business calculus.
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