What is the product rule formula?

(fg)'=f'g+fg': the derivative of the first times the second, plus the first times the derivative of the second.

What is the quotient rule formula?

(frac fg)'=(f'g-fg')/(g^2): bottom times the derivative of the top, minus top times the derivative of the bottom, all over the bottom squared.

Does the order matter in the product rule?

No, because addition doesn't depend on order: f'g+fg' equals fg'+f'g. In the quotient rule the order does matter, because of the subtraction.

Can you use the product rule instead of the quotient rule?

Yes: write frac fg=f· g^-1 and use the product rule with the chain rule on g^-1. It gives the same answer, sometimes in a less tidy form.

Do I need to simplify after using the quotient rule?

Simplify the top (multiply out and combine like terms) and leave the bottom as a square. Factoring the top helps if you need to find where the derivative is zero.

Product Rule and Quotient Rule: How and When to Use Each

Why isn't the derivative of a product the product of the derivatives?

Sums can be differentiated term by term, so it's tempting to think products work the same way. A quick test shows they don't. Take $x \cdot x$ , which is $x^{2}$ , so its derivative is $2 x$ by the power rule. The derivatives of the two factors are $1$ and $1$ , and their product is $1$ , not $2 x$ . So a product needs its own rule.

A picture explains the real answer. Think of $f (x) g (x)$ as the area of a rectangle with sides $f$ and $g$ . When $x$ changes a little, both sides change, and the area grows along both edges: by about (change in $f$ ) times $g$ along one edge, plus $f$ times (change in $g$ ) along the other. That's the product rule.

What is the product rule?

The product rule

\frac{d}{d x} [f (x) g (x)] =

f^{'} (x) g (x) + f (x) g^{'} (x)

Why it works

Start from the definition: the difference quotient has $f (x + h) g (x + h) - f (x) g (x)$ on top and $h$ on the bottom.
Subtract and add $f (x + h) g (x)$ on top, which adds zero. The top becomes $f (x + h) g (x + h) - f (x + h) g (x)$ $+ f (x + h) g (x) - f (x) g (x)$ .
Group in pairs: $f (x + h) \cdot \frac{g (x + h) - g (x)}{h} +$ $g (x) \cdot \frac{f (x + h) - f (x)}{h}$ .
As $h \to 0$ , the fractions become $g^{'} (x)$ and $f^{'} (x)$ , and $f (x + h) \to f (x)$ because a differentiable function is continuous. The total becomes $f (x) g^{'} (x) + g (x) f^{'} (x)$ .

In words: differentiate the first factor and leave the second alone, then leave the first alone and differentiate the second, and add.

Worked example: Product rule, checked by multiplying out

Differentiate $y = (x^{2} + 1) (3 x - 5)$ .

Name the factors: $f = x^{2} + 1$ , so $f^{'} = 2 x$ ; and $g = 3 x - 5$ , so $g^{'} = 3$ .
Product rule: $y^{'} =$ $f^{'} g + f g^{'} =$ $2 x (3 x - 5) + (x^{2} + 1) \cdot 3$ .
Multiply out: $6 x^{2} - 10 x + 3 x^{2} + 3$ .
Combine like terms: $9 x^{2} - 10 x + 3$ .
Check: multiplying out first gives $y = 3 x^{3} - 5 x^{2} + 3 x - 5$ , and the power rule gives $y^{'} = 9 x^{2} - 10 x + 3$ . ✓

Answer: $y^{'} = 9 x^{2} - 10 x + 3$

Worked example: A product that can't be multiplied out

Differentiate $y = x^{2} \cos x$ .

$f = x^{2}$ , so $f^{'} = 2 x$ ; $g = \cos x$ , so $g^{'} = - \sin x$ .
Product rule: $y^{'} =$ $2 x \cos x + x^{2} (- \sin x)$ .
Simplify the sign: $y^{'} =$ $2 x \cos x - x^{2} \sin x$ .

Answer: $y^{'} =$ $2 x \cos x - x^{2} \sin x$

Common mistake

Using the product rule on a number times a function. In $5 \sin x$ , the $5$ is a constant multiple: the derivative is simply $5 \cos x$ . (The product rule gives the same answer, because the derivative of $5$ is $0$ , but it's extra work.)

What is the quotient rule?

The quotient rule

\frac{d}{d x} [\frac{f (x)}{g (x)}] =

\frac{f^{'} (x) g (x) - f (x) g^{'} (x)}{g (x)^{2}}

, wherever

g (x) \neq 0

Why it works

Call the quotient $Q = \frac{f}{g}$ . Then $f = Q g$ , a product.
Product rule on $f = Q g$ : $f^{'} = Q^{'} g + Q g^{'}$ .
Solve for $Q^{'}$ : $Q^{'} g = f^{'} - Q g^{'}$ , so $Q^{'} =$ $\frac{f^{'} - Q g^{'}}{g}$ .
Replace $Q$ by $\frac{f}{g}$ and multiply top and bottom by $g$ : $Q^{'} =$ $\frac{f^{'} g - f g^{'}}{g^{2}}$ .

A common way to remember it, writing "high" for the top and "low" for the bottom: low times d-high, minus high times d-low, over low squared. Starting with the bottom function is what makes the sign come out right.

Worked example: The quotient rule

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

Top $f = 2 x + 1$ , so $f^{'} = 2$ . Bottom $g = x^{2} + 3$ , so $g^{'} = 2 x$ .
Quotient rule: $y^{'} =$ $\frac{2 (x^{2} + 3) - (2 x + 1) (2 x)}{(x^{2} + 3)^{2}}$ .
Multiply out the top: $2 x^{2} + 6 - (4 x^{2} + 2 x)$ .
Simplify: $2 x^{2} + 6 - 4 x^{2} - 2 x =$ $- 2 x^{2} - 2 x + 6$ .
Leave the bottom squared: $y^{'} =$ $\frac{- 2 x^{2} - 2 x + 6}{(x^{2} + 3)^{2}}$ .

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

Worked example: Where the derivative of tan x comes from

Use the quotient rule to differentiate $\tan x =$ $\frac{\sin x}{\cos x}$ .

$f = \sin x$ , $f^{'} = \cos x$ ; $g = \cos x$ , $g^{'} = - \sin x$ .
Quotient rule: $\frac{\cos x \cdot \cos x - \sin x \cdot (- \sin x)}{\cos^{2} x} =$ $\frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x}$ .
Since $\sin^{2} x + \cos^{2} x = 1$ , this is $\frac{1}{\cos^{2} x}$ , which is written $\sec^{2} x$ .

Answer: $\frac{d}{d x} \tan x =$ $\sec^{2} x$

Common mistake

Swapping the order on top. $f^{'} g - f g^{'}$ and $f g^{'} - f^{'} g$ differ by a minus sign, so starting with the wrong term flips the sign of the whole answer. Also, keep the $g^{2}$ on the bottom: it's easy to forget.

When should you avoid the quotient rule?

The quotient rule always works for a quotient, but it isn't always the fastest or safest route. Look at the bottom first:

If the bottom is	Do this instead	Example
a number	take it out as a constant multiple	$\frac{x^{3} + 2 x}{4} =$ $\frac{1}{4} (x^{3} + 2 x)$ , derivative $\frac{1}{4} (3 x^{2} + 2)$
a single term	split into separate powers	$\frac{x^{3} + 2 x}{x^{2}} =$ $x + 2 x^{- 1}$ , derivative $1 - 2 x^{- 2}$
a sum or difference	use the quotient rule	$\frac{4}{x^{3} + 2 x}$

How do you use the product rule with three factors?

Group two of the factors and use the product rule twice. The result has a simple pattern: one term for each factor, with that factor differentiated and the other two left alone.

(f g h)^{'} = f^{'} g h + f g^{'} h + f g h^{'}

Worked example: Three factors

Differentiate $y = x e^{x} \sin x$ .

The three factors are $x$ (derivative $1$ ), $e^{x}$ (derivative $e^{x}$ ) and $\sin x$ (derivative $\cos x$ ).
One term per factor: $y^{'} =$ $1 \cdot e^{x} \sin x + x \cdot e^{x} \cdot \sin x +$ $x \cdot e^{x} \cdot \cos x$ .
Factor out $e^{x}$ : $y^{'} =$ $e^{x} (\sin x + x \sin x + x \cos x)$ .

Answer: $y^{'} =$ $e^{x} (\sin x + x \sin x + x \cos x)$

How do the product and quotient rules combine with the chain rule?

In larger problems, the rules nest. To differentiate $x^{2} \sqrt{x^{2} + 1}$ , the last operation is multiplying, so the product rule comes first; differentiating the factor $\sqrt{x^{2} + 1}$ then needs the chain rule. Find the outermost operation, apply its rule, and repeat on each piece.

The product rule and quotient rule: how and when to use each