Difficulty: Intermediate | Prerequisites: Limit definition of the derivative, derivatives of basic functions
Knowing individual derivative formulas is only half the picture. Nearly every function you encounter in practice is a combination of simpler functions: products, quotients, compositions, or implicitly defined relationships. This set of notes covers the rules for pulling those combinations apart. The chain rule, in particular, is the single most-used technique in all of differential calculus and reappears throughout Calc II in integration by substitution, parametric differentiation, and series work.
The product rule, quotient rule, and chain rule let you differentiate any combination of basic functions. Implicit differentiation handles equations where y is not isolated. Logarithmic differentiation simplifies ugly products and variable-exponent functions. Parametric curves use dy/dt divided by dx/dt.
Constant multiple rule
Constants pass through the derivative: d/dx[c·u] = c·u'. Think of it as: a constant scaling factor stays put while you differentiate what it multiplies.
Sum/difference rule
The derivative of a sum (or difference) is the sum (or difference) of the derivatives. In simple terms, you can differentiate term by term.
Product rule
The rule for differentiating u·v: d/dx[u·v] = u'·v + u·v'. In simple terms, differentiate the first and leave the second alone, then leave the first alone and differentiate the second, then add.
Quotient rule
The rule for differentiating u/v: d/dx[u/v] = (u'·v - u·v') / v². The mnemonic "ho dee hi minus hi dee ho over ho ho" encodes the structure.
Chain rule
The rule for differentiating compositions f(g(x)): take the derivative of the outside function evaluated at the inside, then multiply by the derivative of the inside. The most frequently used differentiation rule.
Implicit differentiation
A technique for finding dy/dx when y is defined implicitly by an equation (e.g. x² + y² = 1) rather than explicitly as y = f(x). You differentiate both sides with respect to x, treating y as a function of x.
Logarithmic differentiation
A technique where you take the natural log of both sides of y = f(x), simplify using log properties, then differentiate implicitly. Useful for functions with variable exponents or complicated products.
Parametric differentiation
Finding dy/dx when x and y are both given as functions of a parameter t. The formula is (dy/dt) / (dx/dt).
Constant multiple rule:
\frac{d}{dx}[c \cdot u] = c \cdot u'Pull constants out, then differentiate. Example: d/dx[7x³] = 7·3x² = 21x².
Sum/difference rule:
\frac{d}{dx}[u \pm v] = u' \pm v'Differentiate each term separately. Example: d/dx[x² + sin(x)] = 2x + cos(x).
These two rules together mean you can handle any polynomial or any sum of basic functions term by term, pulling out constants as you go. They are so natural that students often use them without thinking, which is fine, but be careful not to extend the same logic to products and quotients.
Product rule:
\frac{d}{dx}[u \cdot v] = u' \cdot v + u \cdot v'You cannot just multiply the derivatives together. d/dx[u·v] ≠ u'·v'. The product rule says: differentiate the first factor while leaving the second alone, then leave the first alone and differentiate the second, and add the results.
For products of three functions:
\frac{d}{dx}[u \cdot v \cdot w] = u' \cdot v \cdot w + u \cdot v' \cdot w + u \cdot v \cdot w'The pattern extends naturally: differentiate one factor at a time, leaving the others untouched, and sum all the terms.
Quotient rule:
\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u' \cdot v - u \cdot v'}{v^2}You cannot just divide the derivatives. The mnemonic "ho dee hi minus hi dee ho over ho ho" helps: bottom times derivative of top, minus top times derivative of bottom, all over bottom squared.
Note: many quotient rule problems can also be solved by rewriting the quotient as a product (u·v⁻¹) and using the product rule with the chain rule. Some students find this approach less error-prone.
\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)In words: derivative of the outside (evaluated at the inside) times derivative of the inside.
This is the most important differentiation rule. Any time you see a function inside another function, the chain rule applies. Examples:
d/dx[sin(3x)] = cos(3x)·3 = 3cos(3x). The outside is sin, the inside is 3x.
d/dx[(x² + 1)⁵] = 5(x² + 1)⁴·2x = 10x(x² + 1)⁴. The outside is the fifth power, the inside is x² + 1.
d/dx[e^(x²)] = e^(x²)·2x. The outside is eˣ, the inside is x².
The chain rule can be applied multiple times for deeply nested compositions. For f(g(h(x))):
\frac{d}{dx}[f(g(h(x)))] = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)Each layer peels off one "outside" derivative and passes control inward.
Used when y is defined implicitly by an equation like x² + y² = 25, rather than explicitly as y = f(x).
The procedure has three steps:
Differentiate both sides of the equation with respect to x. Every time you differentiate a term involving y, attach a dy/dx factor (this is the chain rule at work, since y is a function of x).
Collect all terms containing dy/dx on one side of the equation.
Factor out dy/dx and solve for it:
Move all terms with dy/dx to the left-hand side
Factor dy/dx out
Divide both sides by whatever is left
Worked example: x² + y² = 25
Differentiating: 2x + 2y·(dy/dx) = 0
Solving: 2y·(dy/dx) = -2x, so dy/dx = -x/y.
The answer is in terms of both x and y, which is typical for implicit differentiation. You can substitute a specific point (x₀, y₀) on the curve to get a numerical slope.
Tangents to parametric curves
Given x = f(t), y = g(t), the slope of the curve at a point is:
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \quad \text{provided } \frac{dx}{dt} \neq 0Differentiate each coordinate with respect to the parameter t, then divide. This comes up in Calc II when working with parametric equations and polar coordinates.
Logarithmic differentiation (technique 1, the main one)
For complicated products, quotients, or functions with variable exponents (like x^x), the steps are:
Take the natural log of both sides: ln(y) = ln(f(x))
Use log properties (log of products becomes sums, log of quotients becomes differences, log of powers brings exponents down) to expand the right-hand side
Differentiate both sides. The left-hand side becomes (1/y)·(dy/dx) by implicit differentiation
Solve for dy/dx
Substitute f(x) back in for y so the final answer is in terms of x only
Logarithmic differentiation (technique 2, simpler case)
When the function is already y = ln(something complicated), just expand using log properties and differentiate directly. No implicit differentiation step needed, no substitution at the end.
Students often multiply derivatives together for products: d/dx[u·v] ≠ u'·v'. This is the single most common differentiation error. Use the product rule.
Similarly, dividing derivatives for quotients: d/dx[u/v] ≠ u'/v'. Use the quotient rule.
Forgetting the "inside derivative" when applying the chain rule. d/dx[sin(3x)] = cos(3x)·3, not just cos(3x). Every chain rule application must include the derivative of the inner function.
In implicit differentiation, forgetting to attach dy/dx when differentiating terms that contain y. If you differentiate y³ with respect to x, the result is 3y²·(dy/dx), not 3y².
⚠️ Product rule and chain rule problems dominate exams. Expect at least one problem requiring both in the same computation.
⚠️ Implicit differentiation is tested as its own question type. Be ready to find dy/dx and evaluate it at a given point.
⚠️ Logarithmic differentiation is the standard approach for functions like y = x^(sin x) or y = (x²·√(x+1)) / (x³ + 2)⁴. If you see variable exponents or a product/quotient of many terms, reach for the log.
⚠️ The quotient rule mnemonic ("ho dee hi minus hi dee ho over ho ho") is worth memorising, but also practise rewriting quotients as products to build flexibility.
d/dx[x²·sin(x)] = ? (Which rule?)
d/dx[e^(5x)] = ? (Which rule?)
If x² + y³ = 10, what is dy/dx?
True or false: d/dx[u/v] = u'/v'.
For x = t², y = t³, what is dy/dx in terms of t?
Answers: 1. 2x·sin(x) + x²·cos(x), product rule. 2. 5e^(5x), chain rule. 3. dy/dx = -2x/(3y²). 4. False. 5. (3t²)/(2t) = 3t/2.
Q: Find d/dx[(x³ + 2)⁷].
A: By the chain rule: 7(x³ + 2)⁶ · 3x² = 21x²(x³ + 2)⁶.
Q: Differentiate f(x) = x²·eˣ.
A: Product rule: f'(x) = 2x·eˣ + x²·eˣ = eˣ(2x + x²) = x·eˣ(2 + x).
Q: Use implicit differentiation to find dy/dx if x·y + y² = 3.
A: Differentiate: y + x·(dy/dx) + 2y·(dy/dx) = 0. Factor: (dy/dx)(x + 2y) = -y. So dy/dx = -y/(x + 2y).
Q: Use logarithmic differentiation to find dy/dx for y = x^x.
A: ln(y) = x·ln(x). Differentiate: (1/y)·(dy/dx) = ln(x) + 1. So dy/dx = y·(ln(x) + 1) = x^x·(ln(x) + 1).
Q: Given x = cos(t), y = sin(t), find dy/dx.
A: dy/dx = (dy/dt)/(dx/dt) = cos(t)/(-sin(t)) = -cot(t).
The chain rule is the theoretical basis for integration by substitution (u-substitution) in Calc II: substitution reverses the chain rule. Implicit differentiation returns in related rates problems and when working with inverse functions. Parametric differentiation extends to arc length and surface area calculations for parametric and polar curves.
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