Difficulty: Intermediate | Prerequisites: Limits, continuity, and basic algebra
Differentiation is the central operation of calculus. Once you know how to take a derivative, you can find instantaneous rates of change, locate maxima and minima, sketch curves, and solve applied business problems. This set of notes covers the core rules: power, product, quotient, and chain, plus the derivatives of logarithmic and exponential functions. These are mechanical skills, and fluency with them is assumed on every exam question that follows.
The power rule handles polynomials. The product and quotient rules handle products and fractions of functions. The chain rule handles compositions (a function inside a function). The derivatives of ln(x) and eˣ have their own patterns, and they combine with the chain rule constantly.
Derivative
The instantaneous rate of change of a function at a given point. Notated as f'(x), y', or dy/dx.
Think of it as: how fast the output is changing right now, at this exact input.
Power rule
If f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. The exponent comes down as a coefficient, and the new exponent is one less.
In simple terms, multiply by the power and reduce the power by one.
Product rule
The rule for differentiating a product of two functions: (fg)' = f'g + fg'.
Think of it as: derivative of the first times the second, plus the first times the derivative of the second.
Quotient rule
The rule for differentiating a fraction of two functions: (f/g)' = (f'g − fg')/g².
In simple terms, low d-high minus high d-low, all over the square of what is below.
Chain rule
The rule for differentiating a composition f(g(x)): the derivative is f'(g(x)) · g'(x). Differentiate the outer function, leave the inner function alone, then multiply by the derivative of the inner function.
Think of it as: peel the layers, one at a time, and multiply.
Derivative of ln(x)
d/dx[ln(x)] = 1/x. With the chain rule: d/dx[ln(u)] = u'/u.
Derivative of eˣ
d/dx[eˣ] = eˣ. With the chain rule: d/dx[eᵘ] = eᵘ · u'.
For y = 7x⁵ − 10x + 8, apply the power rule term by term: y' = 35x⁴ − 10. The constant 8 vanishes.
Constants always differentiate to zero. The coefficient of x (here, −10) stays as is because x¹ becomes x⁰ = 1.
For y = 4·ln(x⁶), first simplify using the log rule: 4·ln(x⁶) = 24·ln(x). Then differentiate: y' = 24/x.
Alternatively, use the chain rule directly on ln(x⁶): derivative of ln(u) is u'/u, where u = x⁶. So y' = 4 · (6x⁵/x⁶) = 4 · (6/x) = 24/x.
Both approaches give the same result. Simplifying first is usually faster.
For y = x³·e²ˣ, use the product rule: y' = (x³)'·e²ˣ + x³·(e²ˣ)'.
(x³)' = 3x². For (e²ˣ)', apply the chain rule: e²ˣ · 2 = 2e²ˣ.
So y' = 3x²·e²ˣ + x³·2e²ˣ = e²ˣ(3x² + 2x³) = x²·e²ˣ·(3 + 2x).
Always factor completely on the exam. Pull out common factors (here, x² and e²ˣ).
For y = e^(x² + 5x + 4), the outer function is eᵘ and the inner function is u = x² + 5x + 4.
Derivative of eᵘ is eᵘ · u'. Here u' = 2x + 5.
So y' = (2x + 5)·e^(x² + 5x + 4).
For y = e⁴ˣ/x², use the quotient rule: y' = [(e⁴ˣ)'·x² − e⁴ˣ·(x²)'] / (x²)².
(e⁴ˣ)' = 4e⁴ˣ (chain rule). (x²)' = 2x.
y' = [4e⁴ˣ·x² − e⁴ˣ·2x] / x⁴ = e⁴ˣ·x·(4x − 2) / x⁴ = e⁴ˣ·(4x − 2) / x³.
Factor out 2: y' = 2e⁴ˣ·(2x − 1) / x³.
For y = e^(3x²), first derivative: y' = 6x·e^(3x²) (chain rule).
Second derivative: apply the product rule to 6x·e^(3x²). y'' = 6·e^(3x²) + 6x·(6x·e^(3x²)) = 6e^(3x²)(1 + 6x²).
Power rule: d/dx[xⁿ] = nxⁿ⁻¹
Constant rule: d/dx[c] = 0
Sum/difference rule: d/dx[f ± g] = f' ± g'
Product rule: d/dx[f·g] = f'g + fg'
Quotient rule: d/dx[f/g] = (f'g − fg')/g²
Chain rule: d/dx[f(g(x))] = f'(g(x))·g'(x)
Natural log: d/dx[ln(x)] = 1/x
General log chain: d/dx[ln(u)] = u'/u
Exponential: d/dx[eˣ] = eˣ
General exponential chain: d/dx[eᵘ] = eᵘ·u'
Log simplification: ln(xⁿ) = n·ln(x) (useful before differentiating)
Students often forget that the derivative of a constant is zero. If you see "+8" in a polynomial, it disappears after differentiation.
When using the product rule, students sometimes multiply the two derivatives together instead of applying f'g + fg'. The product rule is a sum, not a product of derivatives.
A very common chain rule error is forgetting to multiply by the derivative of the inner function. If you differentiate e^(3x) and get just e^(3x) without the factor of 3, you have missed the chain rule.
Students often apply the quotient rule when simplification would make it unnecessary. For example, e⁴ˣ/x² can be handled with the quotient rule, but first simplifying ln(x⁶) to 6·ln(x) avoids the chain rule entirely.
⚠️ The power rule appears on nearly every exam question. Make it automatic.
⚠️ Product rule combined with exponential functions (like x³·e²ˣ) is a favourite exam pattern. Always factor the result completely.
⚠️ Chain rule with e^(something) is heavily tested. The derivative of e^(u) is always e^(u)·u'.
⚠️ Simplifying ln(xⁿ) to n·ln(x) before differentiating saves time and reduces errors. Examiners expect you to recognise this shortcut.
⚠️ Second derivatives require applying the product rule to the first derivative when it is itself a product.
Derivatives tell you how quickly a quantity is changing at a given moment. In business, this translates directly to marginal cost, marginal revenue, and marginal profit, which are all derivatives of their respective functions.
True or False: The derivative of 5x³ + 2 is 15x² + 2.
Fill in the blank: d/dx[ln(x⁴)] = ___.
True or False: The derivative of e^(5x) is e^(5x).
Fill in the blank: Using the product rule on x²·eˣ gives ___.
Q: Differentiate y = 7x⁵ − 10x + 8.
A: y' = 35x⁴ − 10. The constant 8 drops off, and the power rule gives 5·7 = 35 on the first term.
Q: Differentiate y = 4·ln(x⁶).
A: Simplify to 24·ln(x), then differentiate: y' = 24/x.
Q: Find y' for y = x³·e²ˣ and factor completely.
A: Product rule: y' = 3x²·e²ˣ + x³·2e²ˣ = x²·e²ˣ·(3 + 2x).
Q: Differentiate y = e^(x² + 5x + 4).
A: Chain rule: y' = (2x + 5)·e^(x² + 5x + 4).
Q: Differentiate y = e⁴ˣ/x² and factor.
A: Quotient rule then factor: y' = 2e⁴ˣ·(2x − 1)/x³.
Q: Find the second derivative of y = e^(3x²).
A: First derivative: y' = 6x·e^(3x²). Second derivative (product rule): y'' = 6e^(3x²)·(1 + 6x²).
Every differentiation rule here feeds directly into applied problems: marginal cost/revenue/profit, tangent line equations, curve sketching, and optimisation. The chain rule in particular is the bridge between simple derivatives and the exponential growth and decay models used in business and science.
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