Difficulty: Intermediate | Prerequisites: Basic derivative formulas (power rule, trig derivatives, exponential and log derivatives)
The product rule, chain rule, and quotient rule let you differentiate combinations of functions: products, compositions, and quotients. Most exam problems require two or more of these rules in a single problem. Mastering the mechanics and knowing which rule to apply first is the core skill here.
Product rule
The derivative of a product of two functions: (fg)' = f'g + fg'.
In simple terms, differentiate the first and multiply by the second, then add the first multiplied by the derivative of the second.
Chain rule
The derivative of a composite function: if y = f(g(x)), then y' = f'(g(x)) times g'(x).
Think of it as peeling layers: differentiate the outer function (leaving the inner function alone), then multiply by the derivative of the inner function.
Quotient rule
The derivative of a quotient f/g: (f/g)' = (f'g - fg') / g^2.
In simple terms, "low d-high minus high d-low, over low squared."
Inverse trigonometric derivatives
The derivative of arctan(x) is 1/(1 + x^2). The derivative of arcsin(x) is 1/sqrt(1 - x^2).
These come up whenever you see tan^(-1) or sin^(-1) in a function you need to differentiate.
Exponential derivative (base other than e)
The derivative of a^x is a^x times ln(a). This follows from rewriting a^x as e^(x ln a) and applying the chain rule.
In simple terms, exponentials with base a pick up a factor of ln(a) when differentiated.
Logarithmic derivative
The derivative of ln(x) is 1/x. Combined with the chain rule, the derivative of ln(g(x)) is g'(x)/g(x).
Product rule with trig functions
Example: f(x) = 7x cos(3x) - 6 sin(3x).
Apply the product rule to 7x cos(3x): derivative is 7 cos(3x) + 7x(-sin(3x))(3) = 7 cos(3x) - 21x sin(3x).
The derivative of -6 sin(3x) is -6(3) cos(3x) = -18 cos(3x).
Combining: f'(x) = 7 cos(3x) - 21x sin(3x) - 18 cos(3x) = -21x sin(3x) - 11 cos(3x).
The chain rule appears inside the product rule whenever the argument is not just x (here it is 3x, producing the extra factor of 3).
Product rule with exponentials and square roots
Example: f(x) = e^(-3x)(6x + 1)^(1/2).
Treat this as a product of e^(-3x) and (6x + 1)^(1/2).
Derivative of e^(-3x) is -3e^(-3x). Derivative of (6x + 1)^(1/2) is (1/2)(6x + 1)^(-1/2)(6) = 3/(6x + 1)^(1/2).
Product rule gives: 3e^(-3x)/(6x + 1)^(1/2) - 3e^(-3x)(6x + 1)^(1/2).
Factor out 3e^(-3x)/(6x + 1)^(1/2) to get (3e^(-3x)/(6x + 1)^(1/2))(1 - (6x + 1)) = -18xe^(-3x)/(6x + 1)^(1/2).
Chain rule with nested functions
Example: f(x) = sin(cos(x)).
The outer function is sin(u) where u = cos(x). The derivative of sin(u) is cos(u), and u' = -sin(x).
Result: f'(x) = cos(cos(x)) times (-sin(x)) = -sin(x) cos(cos(x)).
Example: f(x) = sqrt(x^2 - 6x).
Rewrite as (x^2 - 6x)^(1/2). The chain rule gives (1/2)(x^2 - 6x)^(-1/2) times (2x - 6).
Simplify: f'(x) = (x - 3)/sqrt(x^2 - 6x).
Quotient rule with logarithms
Example: f(t) = 2 ln(t) / (5 - ln(t)).
Numerator derivative: 2/t. Denominator derivative: -1/t.
Quotient rule: f'(t) = [(2/t)(5 - ln t) - (2 ln t)(-1/t)] / (5 - ln t)^2.
Simplify the numerator: (2/t)(5 - ln t) + (2 ln t)/t = (10 - 2 ln t + 2 ln t)/t = 10/t.
Result: f'(t) = 10 / [t(5 - ln t)^2].
Example: y = ln(x) / (8 + x).
Numerator derivative: 1/x. Denominator derivative: 1.
Quotient rule: dy/dx = [(1/x)(8 + x) - ln(x)(1)] / (8 + x)^2.
Simplify: dy/dx = (8 + x - x ln x) / [x(8 + x)^2].
The factor of 1/x from the numerator derivative carries through to the final denominator.
Exponentials with base other than e
To differentiate a^x, rewrite as e^(x ln a). By the chain rule, the derivative is e^(x ln a) times ln(a) = a^x ln(a).
Example: f(x) = 3 times 4^x - 5x^4 ln(4).
The derivative of 3 times 4^x is 3 times 4^x times ln(4). The derivative of -5x^4 ln(4) is -20x^3 ln(4).
Result: f'(x) = (3 times 4^x - 20x^3) ln(4).
Notice that ln(4) factors out of both terms.
Inverse trigonometric functions combined with logarithms
Example: f(x) = 3 arctan(x/2) + 2 ln(4 + x^2).
Derivative of 3 arctan(x/2): chain rule gives 3 times 1/(1 + (x/2)^2) times (1/2) = 3 times 1/(1 + x^2/4) times (1/2) = 6/(4 + x^2).
Derivative of 2 ln(4 + x^2): chain rule gives 2 times (2x)/(4 + x^2) = 4x/(4 + x^2).
Both terms share the denominator 4 + x^2. Adding: f'(x) = (6 + 4x)/(4 + x^2) = 2(3 + 2x)/(4 + x^2).
Combining fractions with a common denominator is a standard simplification step on exams.
(fg)' = f'g + fg' \quad \text{(Product Rule)}\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \quad \text{(Quotient Rule)}\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \quad \text{(Chain Rule)}\frac{d}{dx} \ln x = \frac{1}{x}, \quad \frac{d}{dx} e^x = e^x\frac{d}{dx} a^x = a^x \ln a\frac{d}{dx} \sin x = \cos x, \quad \frac{d}{dx} \cos x = -\sin x\frac{d}{dx} \tan^{-1} x = \frac{1}{1 + x^2}\frac{d}{dx} \sqrt{u} = \frac{u'}{2\sqrt{u}} \quad \text{(Chain Rule applied)}Students often forget the chain rule factor when differentiating composite functions. If the argument is 3x rather than x, the derivative picks up a factor of 3. Missing this is the single most common error.
When using the quotient rule, students sometimes get the sign wrong in the numerator. It is f'g minus fg', not the other way round. "Low d-high minus high d-low" is worth memorising.
Students sometimes treat ln(x) as though its derivative is 1, rather than 1/x. Similarly, the derivative of ln(g(x)) is g'(x)/g(x), not just g'(x).
When differentiating a^x (base other than e), students often write the derivative as x times a^(x-1) (the power rule pattern). That is wrong. The correct form is a^x ln(a).
Problems that combine two or more rules (product + chain, quotient + chain) are standard. Expect to use at least two rules per problem.
Simplification after applying the rules is often required. Factoring common terms, combining fractions over a common denominator, and cancelling are all tested.
Derivatives of inverse trig functions (especially arctan) combined with logarithms appear frequently. Look for the opportunity to combine terms over a shared denominator.
Derivatives of a^x (base other than e) with a ln(a) factor: examiners like testing whether students confuse the exponential rule with the power rule.
True or false: The derivative of 5^x is x times 5^(x-1). Answer: False. The derivative is 5^x ln(5).
Fill in the blank: The derivative of sin(cos(x)) is ______. Answer: -sin(x) cos(cos(x)).
True or false: The quotient rule gives (f'g + fg')/g^2. Answer: False. The numerator is f'g - fg' (subtraction, not addition).
Fill in the blank: The derivative of arctan(x) is ______. Answer: 1/(1 + x^2).
True or false: The derivative of ln(3x) is 1/(3x). Answer: False. By the chain rule, it is 3/(3x) = 1/x.
Q: Find the derivative of f(x) = 7x cos(3x) - 6 sin(3x).
A: f'(x) = -21x sin(3x) - 11 cos(3x). Apply the product rule to the first term, the chain rule to both, then combine like terms.
Q: Find dy/dx when y = ln(x)/(8 + x).
A: dy/dx = (8 + x - x ln x) / [x(8 + x)^2]. Apply the quotient rule, then simplify by factoring out 1/x from the numerator.
Q: Differentiate f(x) = e^(-3x)(6x + 1)^(1/2).
A: f'(x) = -18xe^(-3x) / (6x + 1)^(1/2). Use the product rule, then factor and simplify.
Q: What is the derivative of f(x) = 3 arctan(x/2) + 2 ln(4 + x^2)?
A: f'(x) = 2(3 + 2x)/(4 + x^2). Differentiate each term using the chain rule, then combine over the common denominator.
Q: Find f'(t) when f(t) = 2 ln(t) / (5 - ln t).
A: f'(t) = 10 / [t(5 - ln t)^2]. Apply the quotient rule and simplify.
These differentiation rules are essential prerequisites for the Fundamental Theorem of Calculus, since FTC1 often requires further differentiation (chain rule on the upper limit, or finding second derivatives of accumulation functions).
The product rule, quotient rule, and chain rule also underpin implicit differentiation and related rates problems, which typically appear later in the course. Fluency with these rules now saves significant time on those topics.
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