Differentiate $y = 4x^3 - 6\sqrt{x} + \frac{5}{x^2}$.

$y' = 12x^2 - \frac{3}{\sqrt{x}} - \frac{10}{x^3}$

Differentiate $y = \frac{3x - 1}{x^2 + 2}$.

$y' = \frac{-3x^2 + 2x + 6}{(x^2 + 2)^2}$

Find the equation of the tangent line to $y = \frac{x^2}{x + 1}$ at $x = 1$.

$y = \frac{3}{4}x - \frac{1}{4}$

Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Power, Product and Quotient Rules

Three rules do most of the differentiating in MCV4U. The power rule handles single terms, the product rule handles two functions multiplied, and the quotient rule handles one divided by another.

How It Works

Power rule. $\frac{d}{dx}x^n = nx^{n - 1}$ for any real exponent $n$ : whole numbers, negatives and fractions. Bring the exponent down and lower it by one. Constants come along for the ride, and a sum is differentiated term by term.

To use it on roots and reciprocals, rewrite them as powers first: $\sqrt{x} = x^{\frac{1}{2}}$ and $\frac{5}{x^2} = 5x^{-2}$ . Then differentiate, and rewrite with positive exponents at the end.

Product rule. If $y = f(x)g(x)$ , then

$y' = f'(x)g(x) + f(x)g'(x)$

Derivative of the first times the second, plus the first times the derivative of the second. The derivative of a product is not the product of the derivatives: $(x \cdot x)' = 2x$ , not $1$ .

Quotient rule. If $y = \frac{f(x)}{g(x)}$ , then

$y' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

The order in the numerator matters because of the minus sign. Start with the derivative of the top. A common memory aid is "low d-high minus high d-low, over low squared".

Which rule? Look at the outermost operation. A plus or minus sign: term by term. Two pieces multiplied, both with $x$ in them: product rule. A fraction with $x$ in the denominator: quotient rule. If the denominator is a single power like $x^3$ , it is faster to split the fraction or rewrite with negative exponents and use the power rule.

Simplify at the end. For a quotient, expand the numerator and collect like terms, but leave the denominator squared and unexpanded. That form shows where the derivative is zero and where it is undefined, which the next units use.

5 Worked Examples

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

Easy
Differentiate $y = 4x^3 - 6\sqrt{x} + \frac{5}{x^2}$ .
Show solution
1. Rewrite as powers: $y = 4x^3 - 6x^{\frac{1}{2}} + 5x^{-2}$ .
2. Power rule on each term: $y' = 12x^2 - 3x^{-\frac{1}{2}} - 10x^{-3}$ .
3. Positive exponents: $y' = 12x^2 - \frac{3}{\sqrt{x}} - \frac{10}{x^3}$ .
Answer: $y' = 12x^2 - \frac{3}{\sqrt{x}} - \frac{10}{x^3}$
Easy
Use the product rule to differentiate $y = (2x + 3)(x^2 - 5)$ .
Show solution
1. $f = 2x + 3$ with $f' = 2$ ; $g = x^2 - 5$ with $g' = 2x$ .
2. $y' = 2(x^2 - 5) + (2x + 3)(2x)$ .
3. Expand and collect: $y' = 2x^2 - 10 + 4x^2 + 6x = 6x^2 + 6x - 10$ .
Answer: $y' = 6x^2 + 6x - 10$
Medium
Differentiate $y = \frac{3x - 1}{x^2 + 2}$ .
Show solution
1. Top $3x - 1$ has derivative $3$ ; bottom $x^2 + 2$ has derivative $2x$ .
2. $y' = \frac{3(x^2 + 2) - (3x - 1)(2x)}{(x^2 + 2)^2}$ .
3. Numerator: $3x^2 + 6 - 6x^2 + 2x = -3x^2 + 2x + 6$ .
Answer: $y' = \frac{-3x^2 + 2x + 6}{(x^2 + 2)^2}$
Medium
Find the equation of the tangent line to $y = \frac{x^2}{x + 1}$ at $x = 1$ .
Show solution
1. Quotient rule: $y' = \frac{2x(x + 1) - x^2}{(x + 1)^2} = \frac{x^2 + 2x}{(x + 1)^2}$ .
2. Slope at $x = 1$ : $\frac{1 + 2}{4} = \frac{3}{4}$ . Point: $y = \frac{1}{2}$ .
3. $y - \frac{1}{2} = \frac{3}{4}(x - 1)$ , so $y = \frac{3}{4}x - \frac{1}{4}$ .
Answer: $y = \frac{3}{4}x - \frac{1}{4}$
Test level
Find the points on $y = \frac{x^2 - 3}{x - 2}$ where the tangent line is horizontal.
Show solution
1. A horizontal tangent means $y' = 0$ .
2. $y' = \frac{2x(x - 2) - (x^2 - 3)}{(x - 2)^2} = \frac{x^2 - 4x + 3}{(x - 2)^2} = \frac{(x - 1)(x - 3)}{(x - 2)^2}$ .
3. A fraction is zero when its numerator is, so $x = 1$ or $x = 3$ . Neither is $2$ , so both are in the domain.
4. $y(1) = \frac{-2}{-1} = 2$ and $y(3) = \frac{6}{1} = 6$ .
Answer: $(1, 2)$ and $(3, 6)$

Learn It on Martian Lab

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

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MCV4U: Power, Product and Quotient Rules

How It Works

5 Worked Examples

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