# The Product and Quotient Rules

Lesson Features »

Lesson Priority: High

Calculus $\longrightarrow$
Discovering Derivatives $\longrightarrow$

Objectives
• Learn the product rule for differentiating a product of two pieces when you know the derivative of each piece
• Learn the quotient rule for differentiating the quotient of two terms when you know the derivative of each term
• Know expert tips and tricks to avoid very common pitfalls for using these techniques mistake-free
• Because the quotient rule is more involved, learn how we can often turn a quotient problem into a multiplication problem, and then proceed with the product rule
Lesson Description

Unlike limit laws, it is not true that the derivative of a product is the product of the derivatives. The derivative of a product has a pattern based approach, though, and we will learn and practice it here, before immediately moving on to how to take the derivative of a quotient.

Practice Problems

Practice problems and worksheet coming soon!

## Unintuitive Derivative Rules

Derivatives are fairly well-behaved in a lot of convenient ways. By now we have studied several derivative laws and rules - some of which were so intuitive that it seemed overkill to state them formally. For example, we learned that the derivative of the sum of terms was equal to the sum of each derivative:$$\frac{d}{dx} \,\, f(x) + g(x) = \frac{d}{dx} \,\, f(x) + \frac{d}{dx} \,\, g(x)$$Hopefully by now this behavior is automatic to you, having used it countless times in practice.Now we turn to slightly more complicated ideas. First, what happens when we need the derivative of a product of functions? Unfortunately, the derivative of a product is NOT equal to the product of each derivative. The reason for this is perhaps beyond your curiosity - what I mean is, most students aren't overly interested in why this is the case. Regardless, all students need to know what to do when asked to find a derivative that involves a product.
Definition:Let $f(x)$ and $g(x)$ be differentiable functions of $x$. It follows that$$\frac{d}{dx} \,\, \bigg( f(x) g(x) \bigg) = \left[\frac{d}{dx} \,\, f(x)\right] \cdot g(x) + f(x) \cdot \left[\frac{d}{dx} \,\, g(x)\right]$$Or, if you prefer prime notation:Let $h(x) = f(x) \cdot g(x)$, where $f$ and $g$ are differentiable functions of $x$. It follows that$$h'(x) = f'(x) \cdot g(x) + f(x)\cdot g'(x)$$
In words, we can say that the derivative of a product of two things is Thing 1 times Thing 2's derivative plus Thing 2 times Thing 1's derivative (for you Dr. Seuss fans out there).

## Using the Product Rule Correctly

On the subject of derivatives of products, you essentially have two jobs: 1) recognize that the problem involves a product and 2) apply the product rule correctly.
You Should Know When you know the derivative of each expression on its own, but you're asked to find the derivative of the product of two expressions, you need to use the Product Rule.
It's very important that you don't take the first job for granted (recognizing the need to use the Product Rule in the first place). It will become less of a concern with practice and exposure, but one of the most common mistakes students make at first is failing to even recognize the fact that the question requires the product rule. Let's examine several situations to better understand when we do and do not need the Product Rule.Examples 1-7For each of the following derivatives, determine whether or not you need to use the Product Rule.

Example 1$$\frac{d}{dx} \,\, xe^x$$$\blacktriangleright$ This example does require the Product Rule, because you know the derivative of $x$, and you know the derivative of $e^x$, and you're being asked to find the derivative of the product of those two expressions.

Example 2$$\frac{d}{dx} \,\, \sin(2x)$$$\blacktriangleright$ This example does not require the use of the Product Rule (though it requires the Chain Rule », which is covered in a near future lesson). The expression is simply not the product of two objects.

Example 3$$\frac{d}{dx} \,\, \big[ x^2 \left(x^2 + 1\right) \big]$$$\blacktriangleright$ Here, we actually could use the Product Rule, but it is much faster to simply multiply it out first, leaving us with a plain polynomial.$$\frac{d}{dx} \,\, \big[ x^2 \left(x^2 + 1\right) \big]$$$$\longrightarrow \frac{d}{dx} \,\, \big[ x^4 + x^2 \big]$$The important takeaway from this example is we not only want to know when we need to use the Product Rule, but also when we can avoid it for the sake of simplicity.

Example 4$$\frac{d}{dx} \,\, x^x$$$\blacktriangleright$ While this one does require a special technique (see logarithmic differentiation »), there is no product here, and the Product Rule therefore is not in play.

Example 5$$\frac{d}{dx} \,\, e^2 \tan(x)$$$\blacktriangleright$ The $e^2$ term is a constant, so there is no need to use the Product Rule. The Constant Rule for derivatives will handle this situation.

Example 6$$\frac{d}{dx} \,\, 2^x \log_2 (x)$$$\blacktriangleright$ There is no shortcut or clever trick to combining this expression into one object. We would take the derivative of this product using the Product Rule.

Example 7$$\frac{d}{dx} \,\, \sin(x) \ln(x)$$$\blacktriangleright$ Here, you need the Product Rule.
Pro Tip When working with the product rule, it is so very important to organize scratch work. The intermediate to advanced questions will require side work, in order to find some of the derivatives you need to plug into the formula. For this reason, you should $\rightarrow$ always $\leftarrow$ set up the formula first before executing it. Put each expression in its place before you actually take any derivatives. It will seem like overkill at first but it will soon be an invaluable habit!
Now let's put the rule into practice! We'll revisit each of the preceding examples above and find the derivative of the ones that required the Product Rule. As mentioned in the Pro Tip above, remember to set up the formula first in your scratch work before taking any derivatives - it's a very important habit!!!Example 1$$\frac{d}{dx} \,\, xe^x$$$\blacktriangleright$ Set things up in line with the formula:$$\frac{d}{dx} \,\, xe^x$$$$= \left( \frac{d}{dx} \,\, x \right)\left(e^x\right) + (x) \left( \frac{d}{dx} \,\, e^x \right)$$Now that we've put everything in the right place, let's take derivatives and finish this problem off.$$= (1) \left(e^x\right) + (x)\left(e^x\right)$$$$= e^x + xe^x$$Example 3$$\frac{d}{dx} \,\, x^2 \left(x^2 + 1\right)$$$\blacktriangleright$ Again, set things up first:$$\frac{d}{dx} \,\, x^2 \left(x^2 + 1\right)$$$$=\left( \frac{d}{dx} \,\, x^2 \right) \left( x^2 + 1\right) + \left(x^2\right) \left( \frac{d}{dx} \,\, x^2 + 1 \right)$$$$=2x \left( x^2 + 1\right) + x^2 (2x)$$$$=2x^3 + 2x + 2x^3 = 4x^3 + 2x$$Recall also that, as pointed out earlier with this example, the best approach would have been to multiply out the expression before taking the derivative, thus avoiding the need for the Product Rule. However we can see by doing it out longform that we got the answer we would have expected.Example 6$$\frac{d}{dx} \,\, 2^x \log_2 (x)$$$$\blacktriangleright \,\,\, =\left(\frac{d}{dx} \,\, 2^x \right) \big(\log_2(x)\big) + \left( 2^x \right) \left(\frac{d}{dx} \,\, \log_2(x) \right)$$$$=\big(2^x \ln(2)\big) \log_2(x) + 2^x \left(\frac{1}{x\ln(2)}\right)$$$$=2^x \ln(2) \log_2 (x) + \frac{2^x}{x\ln(2)}$$Example 7$$\frac{d}{dx} \,\, \sin(x) \ln(x)$$$$\blacktriangleright \,\,\, = \left( \frac{d}{dx} \,\, \sin(x) \right)\big( \ln(x) \big) + \big( \sin(x) \big)\left( \frac{d}{dx} \,\, \ln(x) \right)$$$$=\big( \cos(x) \big) \big( \ln(x) \big) + \big( \sin(x) \big) \left( \frac{1}{x} \right)$$$$=\cos(x) \ln(x) + \frac{\sin(x)}{x}$$
Annoying Pro Tip I just can't stress enough how important it is to take that intermediary step in your pencil and paper calculations and write out each piece and where it belongs before actually taking any derivatives. It seems like overkill now but it will pay huge dividends. As you practice derivatives on some of the more basic expressions like $xe^x$, you'll be sorely tempted to do it all at once - don't give in! It's just a few extra pencil strokes!

## The Quotient Rule

We will now apply the same idea to the derivative of quotients. Similar to products, we must first recognize the situation, and then apply a formulaic approach.
Define: The Quotient Rule for DerivativesLet $f(x)$ and $g(x)$ be differentiable functions of $x$. It follows that$$\frac{d}{dx} \,\, \frac{f(x)}{g(x)} = \frac{g(x) \left(\frac{d}{dx} \,\, f(x)\right) - f(x) \left( \frac{d}{dx} \,\, g(x) \right)}{(\big(g(x)\big)^2}$$Or, if you prefer prime notation:Let $h(x) = \frac{f(x)}{g(x)}$, where $f$ and $g$ are differentiable functions of $x$. It follows that$$h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{\big(g(x)\big)^2}$$
As you can see, the derivative of a product is a little more forgiving, since order didn't matter (for products we just take the derivative of each piece multiplied with the original other piece). With the Quotient Rule, each expression must be in exactly the right place, and order matters because of the subtraction.The biggest mistake students make on exams is forgetting the order of the terms. Here's a memory trick that I used when I first learned this stuff - the following sentence dictates the Quotient Rule for you and helps you keep the order straight:
Define: The Quotient Rule for Derivatives in Words$$\frac{d}{dx} \,\, \frac{f(x)}{g(x)} = \frac{g(x) \left(\frac{d}{dx} \,\, f(x)\right) - f(x) \left( \frac{d}{dx} \,\, g(x) \right)}{(\big(g(x)\big)^2}$$$\longrightarrow$ Low d-high minus high d-low, all over the square of what's below! $\longleftarrow$
"Low" and "below" refer to the denominator expression, and "high" refers to the numerator expression. This is a great exam memory trick, because it's quick and foolproof - if you say it wrong, then the sentence won't rhyme!

Example 8$$\frac{d}{dx} \,\, \frac{x^6}{e^{2x}}$$$\blacktriangleright$ We know the Quotient Rule applies, since we know the derivative of each the numerator and the denominator on their own. Make sure you set up the formula by putting the pieces in their place before we take any derivatives, just like we did for the Product Rule.$$\frac{d}{dx} \,\, \frac{x^6}{e^{2x}}$$$$=\frac{\left(e^{2x} \right)\left(\frac{d}{dx} \,\, x^6 \right) - \left(x^6 \right)\left( \frac{d}{dx} \,\, e^{2x} \right)}{\left(e^{2x}\right)^2}$$$$=\frac{e^{2x}\left(6x^5\right) - \left(x^6\right) \left(2e^{2x} \right)}{e^{4x}}$$Factor out the common $e^{2x}$ in the numerator - this is a very good idea when working on derivatives involving exponentials. Canceling, we are finally left with$$=\frac{6x^5 - 2x^6}{e^{2x}}$$

## Avoiding the Quotient Rule

Since the Quotient Rule is slightly more complicated than the Product Rule, and since we are human, it is preferrable to work with the simpler Product Rule formula when possible. Many quotients can be quickly rewritten as products, through the use of negative exponents. Leveraging that fact, we can turn "Quotient Rule problems" into "Product Rule problems" more often than not.

Example 8 (Revisited)$$\frac{d}{dx} \,\, {x^6}{e^{2x}}$$$\blacktriangleright$ Instead of using the Quotient rule, we can rewrite the expression as a product, which will allow us to use the Product Rule.$$\frac{d}{dx} \,\, \left(x^6\right)\left(e^{-2x}\right)$$$$= x^6 \left(\frac{d}{dx} \,\, e^{-2x} \right) + \left( \frac{d}{dx} \,\, x^6 \right) \left(e^{-2x}\right)$$$$= -2x^6e^{-2x} + 6x^5 e^{-2x}$$$$=\frac{-2x^6+6x^5}{e^{2x}}$$
Pro Tip This trick isn't always going to save you a lot of time or heartache, because taking derivatives of expressions with negative exponents adds more complexity and negative signs into the mix. At some point, a complicated expression is a complicated expression, no matter how you choose to deal with it. This trick tends to work very well with polynomial and exponential expressions. If you have a trig function or a logarithm in the denominator of the thing you're trying to take the derivative of, you're probably better off just using the Quotient Rule verbatim.
Warning! If you are working on a quiz or test with explicit instructions to use the Quotient Rule, then you should not turn the problem into a Product Rule problem. In that case, your instructor would have legitimate reason to dock points off, if they want.

## Put It To The Test

We will be continually practicing both the Product Rule and the Quotient Rule throughout the course of the Calculus journey. Specifically, you will need to practice this a lot in conjunction with other rules - notably the Chain Rule » in the near future. In fact, you may find that your homework on this topic is significantly more difficult than the problems that follow here, if you are working with the Chain Rule simultaneously. You should absolutely work practice problems from the lesson on the Chain Rule if you want to be sure you've mastered anything that can be thrown at you that requires the Product Rule or Quotient Rule.With that in mind, let's dive in.

Example 9Find $\frac{d}{dx} \,\, x^3 e^{-5x}$.
Show solution
$\blacktriangleright$ This problem calls for a straight-forward application of the Product Rule.$$\frac{d}{dx} \,\, x^3 e^{-5x}$$$$= \left( \frac{d}{dx} \,\, x^3 \right)\left( e^{-5x} \right) + \left( x^3 \right)\left( \frac{d}{dx} \,\, e^{-5x} \right)$$$$=\left(3x^2\right)\left(e^{-5x}\right) + \left( x^3 \right) \left( -5e^{-5x} \right)$$$$=e^{-5x} \left[ 3x^2 - 5x^3 \right]$$

Example 10$$\frac{d}{dx} \,\, \sin^2(x)$$
Show solution
$\blacktriangleright$ We will soon learn a direct way to handle this derivative using the Chain Rule », but we can still find the answer with the tools we've learned thus far - namely the Product Rule. Simply write the problem as a product and apply the rule.$$\frac{d}{dx} \,\, \sin(x) \sin(x)$$$$=\left( \frac{d}{dx} \,\, \sin(x) \right) \big( \sin(x) \big) + \left( \frac{d}{dx} \,\, \sin(x) \right) \big( \sin(x) \big)$$$$=\big( \cos(x) \big) \big( \sin(x) \big) + \big( \cos(x) \big) \big( \sin(x) \big)$$$$=2\sin(x) \cos(x)$$

Example 11$$\frac{d}{dx} \,\, \frac{x^2}{2^x}$$
Show solution
$\blacktriangleright$ We know each of these derivatives on their own, so the right way to proceed here is to apply the Quotient Rule.$$\frac{d}{dx} \,\, \frac{x^2}{2^x}$$$$=\frac{\left( 2^x \right)\left( \frac{d}{dx} \,\, x^2 \right) - \left( x^2 \right) \left( \frac{d}{dx} \,\, 2^x \right)}{\left(2^x\right)^2}$$$$=\frac{\left( 2^x \right)\left( 2x \right) - \left( x^2 \right) \left( 2^x \ln(2) \right)}{2^{2x}}$$$$=\cancel{2^x} \left[ \frac{2x - x^2 \ln(2)}{2^{\cancel{2}x}} \right]$$$$=\frac{2x - x^2 \ln(2)}{2^x}$$

Example 12$$\frac{d^2}{dx^2} \,\, \sec(x)$$
Show solution
$\blacktriangleright$ To find the second derivative, we will start by calculating the first derivative.$$\frac{d}{dx} \,\, \sec(x) = \sec(x) \tan(x)$$Take the derivative of this result to find the second derivative. We'll need to employ the product rule.$$\frac{d}{dx} \,\, \sec(x) \tan(x)$$$$=\left( \frac{d}{dx} \,\, \sec(x) \right) \left( \tan(x) \right) + \left( \frac{d}{dx} \,\, \tan(x) \right) \left( \sec(x) \right)$$$$=\left( \sec(x) \tan(x) \right) \left( \tan(x) \right) + \left( \sec^2(x) \right) \left( \sec(x) \right)$$$$= \sec(x) \tan^2 (x) + \sec^3(x)$$

Lesson Takeaways
• Recognize derivatives that require either the Product Rule or Quotient Rule
• For these problems, plug in terms into the appropriate formula first before taking derivatives, to show clear scratch work to both you and your teacher
• Ultimately memorize both the Product Rule and the Quotient Rule
• Use the Product Rule iteratively if needed, e.g. when asked for higher derivatives, and again keep track of your steps with well-organized scratch work
• Practice using both rules for fluency, and practice turning in fully simplified answers at the peril of losing points on tests
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