# Derivatives of Products and Quotient Rule

Quotient Rule Derivative: In this lesson, you will learn the formula for the quotient rule of derivatives. The lesson includes a mnemonic device to help you remember the formula. You will also see two worked-out examples.

The quotient rule formula may be a little difficult to remember. Perhaps a little yodeling-type chant can help you. Imagine a frog yodeling, ‘LO dHI less HI dLO over LO LO.’ In this mnemonic device, LO refers to the denominator function and HI refers to the numerator function.

Let’s translate the frog’s yodel back into the formula for the quotient rule.

LO dHI means denominator times the derivative of the numerator: g(x) times df(x).

less means ‘minus’.

HI dLO means numerator times the derivative of the denominator: f(x) times dg(x).

over means ‘divide by’.

LO LO means to take the denominator times itself: g(x) squared.

## Quotient Derivative Rule

The following problems require the use of the quotient rule. In the following discussion and solutions, the derivative of a function h(x) will be denoted by or h‘(x). The quotient rule is a formal rule for differentiating problems where one function is divided by another. It follows from the limit definition of derivative and is given by .

Remember the rule in the following way. Always start with the “bottom” function and end with the “bottom” function squared. Note that the numerator of the quotient rule is identical to the product rule except that subtraction replaces addition. In the list of problems that follow, most problems are average and a few are somewhat challenging.

## Quotient Rule Derivative Formula

The quotient rule is a formula for taking the derivative of a quotient of two functions. It makes it somewhat easier to keep track of all of the terms. Let’s look at the formula.

If you have function f(x) in the numerator and the function g(x) in the denominator, then the derivative is found using this formula: Take g(x) times the derivative of f(x).In this formula, the d denotes a derivative. So, df(x) means the derivative of function f and dg(x) means the derivative of function g. The formula states that to find the derivative of f(x) divided by g(x), you must:

1. Then from that product, you must subtract the product of f(x) times the derivative of g(x).
2. Finally, you divide those terms by g(x) squared.

## Derivative Product And Quotient Rule

In the previous section, we noted that we had to be careful when differentiating products or quotients. It’s now time to look at products and quotients and see why.

First, let’s take a look at why we have to be careful with products and quotients. Suppose that we have the two functions f(x)=x3 and g(x)=x6. Let’s start by computing the derivative of the product of these two functions. This is easy enough to do directly.

(fg)′=(x3x6)′=(x9)′=9×8

Remember that on occasion we will drop the (x) part on the functions to simplify notation somewhat. We’ve done that in the work above.

Now, let’s try the following.

f′(x)g′(x)=(3×2)(6×5)=18×7

So, we can very quickly see that.

(fg)′≠f′g′

In other words, the derivative of a product is not the product of the derivatives.

Using the same functions we can do the same thing for quotients.

(fg)′=(x3x6)′=(1×3)′=(x−3)′=−3x−4=−3×4

f′(x)

g′(x)=3x26x5=12×3

So, again we can see that,

(fg)′≠f′g′

To differentiate products and we have the Product Rule and the Quotient Rule.

## Partial Derivative Quotient Rule

### Proof from derivative definition and limit properties

Let {\displaystyle f(x)=g(x)/h(x).} Applying the definition of the derivative and properties of limits gives the following proof. ### Proof using implicit differentiation

Let {\displaystyle f(x)={\frac {g(x)}{h(x)}},} so {\displaystyle g(x)=f(x)h(x).} The product rule then gives {\displaystyle g'(x)=f'(x)h(x)+f(x)h'(x).} Solving for {\displaystyle f'(x)} and substituting back for {\displaystyle f(x)} gives: ### Proof using the chain rule

Let {\displaystyle f(x)={\frac {g(x)}{h(x)}}=g(x)h(x)^{-1}.} Then the product rule gives

{\displaystyle f'(x)=g'(x)h(x)^{-1}+g(x)\cdot {\frac {d}{dx}}(h(x)^{-1}).} To evaluate the derivative in the second term, apply the power rule along with the chain rule:

{\displaystyle f'(x)=g'(x)h(x)^{-1}+g(x)\cdot (-1)h(x)^{-2}h'(x).} Finally, rewrite as fractions and combine terms to get

{\displaystyle {\begin{aligned}f'(x)&={\frac {g'(x)}{h(x)}}-{\frac {g(x)
h'(x)}{h(x)^{2}}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}}}

## What is the definition of the quotient rule?

Similar to product rule, the quotient rule is a way of differentiating the quotient, or division of functions. The quotient rule is defined as the quantity of the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all over the denominator squared.

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## How do you find derivatives?

Basically, we can compute the derivative of f(x) using the limit definition of derivatives with the following steps:
1. Find f(x + h).
2. Plug f(x + h), f(x), and h into the limit definition of a derivative.
3. Simplify the difference quotient.
4. Take the limit, as h approaches 0, of the simplified difference quotient.

## What is quotient in division?

The answer after we divide one number by another. dividend ÷ divisor = quotient. Example: in 12 ÷ 3 = 4, 4 is the quotientDivision.