Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Chain Rule

The chain rule differentiates a function inside a function, like $(x^2 - 3x)^7$ or $\sqrt{x^5 - 4x}$ . Expanding those would take all day. The chain rule takes one line.

How It Works

If $y = f(g(x))$ , then $y' = f'(g(x)) \cdot g'(x)$ .

In words: differentiate the outside function and leave the inside alone, then multiply by the derivative of the inside. The lesson's version for powers, which covers most MCV4U questions, is the power of a function rule:

$\frac{d}{dx}[g(x)]^n = n[g(x)]^{n - 1} \cdot g'(x)$

Bring the power down, lower it by one, keep the bracket exactly as it was, and multiply by the derivative of what is inside the bracket.

In Leibniz notation it reads like fractions cancelling. If $y$ depends on $u$ and $u$ depends on $x$ , then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ . That is the version for rates of change: how fast the area grows per second is how fast it grows per centimetre of radius, times how many centimetres per second.

Habits that save marks:

Rewrite roots and reciprocals as powers first. $\sqrt{u} = u^{\frac{1}{2}}$ and $\frac{1}{u^3} = u^{-3}$ .

Never forget the inner derivative. $(3x + 2)^5$ differentiates to $5(3x + 2)^4 \cdot 3$ , and dropping that $3$ is the most common chain rule mistake on any test.

When the chain rule sits inside a product or quotient, do the big rule first and the chain rule for each piece. Then factor out the common brackets, because the answer usually wants a factored form.

Unsure whether you need it? If there is anything other than plain $x$ inside a bracket, a root or a denominator, you need it.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Differentiate $y = (3x + 2)^5$ .
Show solution
1. Outside: the 5th power. Inside: $3x + 2$ , with derivative $3$ .
2. $y' = 5(3x + 2)^4 \cdot 3 = 15(3x + 2)^4$ .
Answer: $y' = 15(3x + 2)^4$
Easy
Differentiate $y = \sqrt{4x^2 + 1}$ .
Show solution
1. Rewrite as a power: $y = (4x^2 + 1)^{\frac{1}{2}}$ .
2. $y' = \frac{1}{2}(4x^2 + 1)^{-\frac{1}{2}} \cdot 8x$ .
3. Simplify: $y' = \frac{4x}{\sqrt{4x^2 + 1}}$ .
Answer: $y' = \frac{4x}{\sqrt{4x^2 + 1}}$
Medium
Differentiate $y = \frac{2}{(x^2 - 5x)^3}$ .
Show solution
1. Rewrite as a power: $y = 2(x^2 - 5x)^{-3}$ .
2. $y' = 2 \cdot (-3)(x^2 - 5x)^{-4} \cdot (2x - 5)$ .
3. Simplify: $y' = \frac{-6(2x - 5)}{(x^2 - 5x)^4}$ .
Answer: $y' = \frac{-6(2x - 5)}{(x^2 - 5x)^4}$
Medium
Differentiate $y = x^2(2x - 1)^4$ and factor your answer.
Show solution
1. It is a product, so product rule first: $y' = 2x(2x - 1)^4 + x^2 \cdot \frac{d}{dx}(2x - 1)^4$ .
2. Chain rule on the second piece: $\frac{d}{dx}(2x - 1)^4 = 4(2x - 1)^3 \cdot 2 = 8(2x - 1)^3$ .
3. So $y' = 2x(2x - 1)^4 + 8x^2(2x - 1)^3$ .
4. Factor out $2x(2x - 1)^3$ : $y' = 2x(2x - 1)^3[(2x - 1) + 4x] = 2x(2x - 1)^3(6x - 1)$ .
Answer: $y' = 2x(2x - 1)^3(6x - 1)$
Test level
Find the equation of the tangent line to $y = \left(\frac{x}{x + 1}\right)^3$ at $x = 1$ .
Show solution
1. Chain rule outside: $y' = 3\left(\frac{x}{x + 1}\right)^2 \cdot \frac{d}{dx}\left(\frac{x}{x + 1}\right)$ .
2. Quotient rule inside: $\frac{(x + 1)(1) - x(1)}{(x + 1)^2} = \frac{1}{(x + 1)^2}$ .
3. Multiply: $y' = \frac{3x^2}{(x + 1)^2} \cdot \frac{1}{(x + 1)^2} = \frac{3x^2}{(x + 1)^4}$ .
4. Slope at $x = 1$ : $\frac{3}{2^4} = \frac{3}{16}$ . Point: $y = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$ .
5. Point-slope: $y - \frac{1}{8} = \frac{3}{16}(x - 1)$ , so $y = \frac{3}{16}x - \frac{1}{16}$ .
Answer: $y = \frac{3}{16}x - \frac{1}{16}$

Learn It on Martian Lab

Martian Lab teaches this in the Chain Rule lesson of its MCV4U Derivatives unit. Watch it, then take the Derivatives practice tests to see if it stuck.

Watch the Chain Rule lesson free

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Free Derivatives practice test, with solutions

More MCV4U topics: Evaluating Limits · Optimization Problems · Related Rates · Curve Sketching Steps · Dot Product and Projections · Cross Product, Area and Torque · Equations of Planes · Intersection of Three Planes

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