Difficulty: Intermediate | Prerequisites: Basic differentiation rules (power, product, quotient, trigonometric derivatives from Ch. 2.3 and 2.4).
The chain rule is the single most-used differentiation technique in calculus. It tells you how to differentiate a composite function, that is, a function built by plugging one function inside another. Every topic from here forward (implicit differentiation, related rates, integration by substitution) leans on it. You should already be comfortable differentiating basic polynomials, trig functions, and using the product and quotient rules before tackling this material.
To differentiate a composite function f(g(x)), multiply the derivative of the outer function (evaluated at the inner) by the derivative of the inner function. In short: "derivative of the outside, times derivative of the inside." Recognising which part is "outer" and which is "inner" is the entire skill.
Composite function
A function formed by applying one function to the output of another, written (f ∘ g)(x) = f(g(x)). Think of it as a two-step machine: x goes into g first, and whatever comes out goes straight into f.
Outer function
The function applied last in a composition. In f(g(x)), f is the outer function. In simple terms, it is the operation you would do second if you were evaluating by hand.
Inner function
The function applied first in a composition. In f(g(x)), g(x) is the inner function. Think of it as the piece "inside the parentheses" that you would evaluate first.
Chain rule
The differentiation rule for composite functions: if y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). In plain English, take the derivative of the outer function, leave the inner function untouched inside it, then multiply by the derivative of the inner function.
Generalised power rule
A special case of the chain rule for functions raised to a power: if y = [u(x)]^n, then dy/dx = n·[u(x)]^(n-1) · u'(x). This is the pattern you will use most often in this assignment.
This is the most common pattern. You have something raised to a power, and the "something" is not just x.
Template: If y = [u(x)]^n, then y' = n·[u(x)]^(n-1) · u'(x).
Example: y = (x^7 + 11)^6. Here the outer function is ( )^6 and the inner function is u = x^7 + 11. So y' = 6(x^7 + 11)^5 · 7x^6 = 42x^6(x^7 + 11)^5.
Example with a square root: f(x) = √(2x^2 + x + 13) = (2x^2 + x + 13)^(1/2). The outer function is ( )^(1/2), the inner is u = 2x^2 + x + 13. So f'(x) = (1/2)(2x^2 + x + 13)^(-1/2) · (4x + 1) = (4x + 1) / [2√(2x^2 + x + 13)].
Example with negative exponents in the inner function: y = (x^3 + 4/x^3)^5 = (x^3 + 4x^(-3))^5. Outer: ( )^5, inner: u = x^3 + 4x^(-3). So dy/dx = 5(x^3 + 4x^(-3))^4 · (3x^2 - 12x^(-4)).
A trig function wrapping something other than plain x, or a trig function itself raised to a power.
Trig of a non-x argument: f(x) = 3cos(x^3). Outer: 3cos( ), inner: u = x^3. So f'(x) = 3·(-sin(x^3))·3x^2 = -9x^2 sin(x^3).
Trig function raised to a power: f(x) = sin^3(x) = [sin(x)]^3. Outer: ( )^3, inner: sin(x). So f'(x) = 3sin^2(x)·cos(x). The tangent line at (0, 0): f'(0) = 3·0·1 = 0, giving y = 0.
sec raised to a power with a fractional argument: f(x) = sec^5(x^(3/4)). This is a double chain: outer ( )^5, middle sec( ), inner x^(3/4). So f'(x) = 5sec^4(x^(3/4)) · sec(x^(3/4))tan(x^(3/4)) · (3/4)x^(-1/4) = (15/4)x^(-1/4) sec^5(x^(3/4)) tan(x^(3/4)).
When a function involves both a fraction structure and a chain rule layer.
Example: y(x) = 6(sin x)^(-3) / x. Rewrite as y = 6x^(-1)(sin x)^(-3). Use the product rule on the two factors, applying the chain rule to (sin x)^(-3). The derivative of (sin x)^(-3) is -3(sin x)^(-4) · cos x. So dy/dx = 6[(-x^(-2))(sin x)^(-3) + x^(-1)(-3(sin x)^(-4) cos x)] = -6(sin x)^(-3)/x^2 - 18 cos x / [x(sin x)^4].
Multiple layers of functions inside each other. Peel from the outside in, multiplying by each inner derivative.
Example: f(x) = cos(√(5/(x+9))). Three layers: cos( ) wraps √( ) wraps 5/(x+9). Work outward to inward: f'(x) = -sin(√(5/(x+9))) · (1/2)(5/(x+9))^(-1/2) · (-5/(x+9)^2). Simplify step by step.
Example: r(t) = sec(√t) · tan(5/t). This requires the product rule first, then the chain rule on each factor. For sec(√t): derivative is sec(√t)tan(√t) · 1/(2√t). For tan(5/t) = tan(5t^(-1)): derivative is sec^2(5/t) · (-5t^(-2)). Then combine via the product rule.
Once you have f'(x) via the chain rule, plug in the x-coordinate to get the slope, then use point-slope form.
Example: f(x) = √(2x^2 + x + 13), find the tangent at (1, 4). We found f'(x) = (4x+1)/[2√(2x^2+x+13)]. At x = 1: f'(1) = 5/(2·4) = 5/8. Tangent line: y - 4 = (5/8)(x - 1), or y = (5/8)x + 27/8.
General chain rule
If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x).
Equivalently, if y = f(u) and u = g(x), then dy/dx = (dy/du) · (du/dx).
Generalised power rule
d/dx [u(x)]^n = n·[u(x)]^(n-1) · u'(x)
Trig-chain formulas
d/dx [sin(u)] = cos(u) · u'
d/dx [cos(u)] = -sin(u) · u'
d/dx [tan(u)] = sec^2(u) · u'
d/dx [sec(u)] = sec(u)tan(u) · u'
d/dx [csc(u)] = -csc(u)cot(u) · u'
d/dx [cot(u)] = -csc^2(u) · u'
In every case, u stands for the inner function and u' is its derivative with respect to x.
The chain rule is how engineers and scientists handle rates of change in systems where one quantity depends on another, which in turn depends on a third. For example, if the temperature inside a reactor depends on pressure, and pressure changes over time, the chain rule connects the rate of temperature change to the rate of pressure change. This same logic appears in economics (marginal cost as production scales), physics (velocity of a particle on a curved path), and machine learning (backpropagation in neural networks is essentially the chain rule applied repeatedly).
Forgetting the inner derivative entirely. Students differentiate the outer function and stop. For y = (x^7 + 11)^6, they write 6(x^7 + 11)^5 and forget to multiply by 7x^6. This is the single most common chain rule error.
Applying the chain rule when it is not needed. If you are differentiating y = x^6, there is no inner function to worry about. The chain rule only applies when you have a composition, not a plain power of x.
Confusing the power of a trig function with the argument. sin^3(x) means [sin(x)]^3, not sin(x^3). These produce completely different derivatives. sin^3(x) differentiates to 3sin^2(x)cos(x). sin(x^3) differentiates to cos(x^3)·3x^2.
Incorrect sign on trig derivatives. When differentiating cos(u), the result is -sin(u)·u', not +sin(u)·u'. The negative sign is easy to drop under pressure.
⚠️ Nearly every derivative problem from this point onward involves the chain rule in some form. Expect it on every exam.
⚠️ Tangent line problems that use the chain rule are a favourite exam format. You need both f'(x) and f(a) to write the equation, so do not skip the evaluation step.
⚠️ Multi-layer compositions (three or more nested functions) appear regularly. The strategy is always the same: peel from the outermost layer inward, multiplying by each inner derivative.
⚠️ The product rule and chain rule often appear together. If the function is a product of two composite pieces, use the product rule first, then the chain rule on each factor separately.
True or False: The derivative of [sin(x)]^4 is 4[sin(x)]^3. (False, you must also multiply by cos(x).)
Fill in the blank: If y = (3x + 1)^10, then y' = ____·(3x + 1)^9 · 3. (Answer: 10)
True or False: d/dx[cos(5x)] = -sin(5x). (False, the answer is -5sin(5x). You need to multiply by the derivative of 5x.)
Fill in the blank: In the composition f(g(x)), the function g(x) is called the ____ function. (Answer: inner)
True or False: The chain rule is needed to differentiate y = x^4 + 7x. (False, this is a simple polynomial with no composition.)
Q: Find the derivative of y = (5x^2 - 3x + 1)^7.
A: y' = 7(5x^2 - 3x + 1)^6 · (10x - 3).
Q: Find the derivative of f(x) = cos^2(4x).
A: Write f(x) = [cos(4x)]^2. Then f'(x) = 2cos(4x) · (-sin(4x)) · 4 = -8cos(4x)sin(4x). This can also be written as -4sin(8x) using the double-angle identity.
Q: Find the derivative of g(t) = √(t^2 + 2t + 5).
A: g(t) = (t^2 + 2t + 5)^(1/2). So g'(t) = (1/2)(t^2 + 2t + 5)^(-1/2) · (2t + 2) = (t + 1)/√(t^2 + 2t + 5).
Q: If y = tan^3(x^2), find dy/dx.
A: Three layers. Outer: ( )^3, middle: tan( ), inner: x^2. dy/dx = 3tan^2(x^2) · sec^2(x^2) · 2x = 6x·tan^2(x^2)·sec^2(x^2).
Q: Find the equation of the tangent line to y = (2x + 1)^4 at x = 0.
A: y(0) = 1^4 = 1. y' = 4(2x+1)^3 · 2 = 8(2x+1)^3. At x = 0: y'(0) = 8. Tangent line: y = 8x + 1.
The chain rule feeds directly into implicit differentiation (Ch. 2.6), where you treat y as a function of x and apply the chain rule whenever you differentiate a term involving y. It also underpins related rates problems (Ch. 3.9), where two quantities each change with time and you connect their rates via the chain rule. Later, in integral calculus, u-substitution is essentially the chain rule run in reverse.
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