# Proofs of the Quotient Rule of Derivatives

The Quotient Rule is one of the most helpful tools in Differential Calculus (or Calculus I) to derive two functions that are being divided. It can be used along with any existing types of functions as long as division operations are present within the given derivation problem.

Here, we will focus mainly on the proofs of the quotient rule formula by applying the concepts of derivation through limits and the chain rule. Also, we will look at some derivation examples of functions that use the quotient rule formula.

##### CALCULUS

Relevant for

Learning about the different proofs of the quotient rule.

See proofs

##### CALCULUS

Relevant for

Learning about the different proofs of the quotient rule.

See proofs

## Proof of The Quotient Rule Using Limits

In this article, you are highly recommended to be familiarized with the topics The Slope of a Tangent Line and Derivatives Using Limits, as a pre-requisite to better understand the proof of the quotient rule using limits.

We can recall that

$$\frac{d}{dx} f(x) = \lim \limits_{h \to 0} {\frac{f(x+h)-f(x)}{h}}$$

Now, we are going to use the following expression:

$$\Upsilon(x) = \frac{f(x)}{g(x)}$$

Then we have,

$$\Upsilon'(x) = \frac{d}{dx} \left(\frac{f(x)}{g(x)}\right)$$

Using limits, we can derive $latex \Upsilon(x)$ by

$$\Upsilon'(x) = \lim \limits_{h \to 0} {\frac{\Upsilon(x+h)-\Upsilon(x)}{h}}$$

By substituting the equation $latex \Upsilon(x) = \frac{f(x)}{g(x)}$, we have

$$\Upsilon'(x) = \lim \limits_{h \to 0} {\frac{{\frac{f(x+h)}{g(x+h)}} – {\frac{f(x)}{g(x)}}}{h}}$$

By getting the least common denominator of the numerator, we have

$$\Upsilon'(x) = \lim \limits_{h \to 0} {\frac{\frac{f(x+h) \cdot g(x) – f(x) \cdot g(x+h)}{g(x+h) \cdot g(x)}}{h}}$$

By applying the rules for fractions, our equation can be re-written as:

$$\Upsilon'(x) = \lim \limits_{h \to 0} {\frac{f(x+h) \cdot g(x) – f(x) \cdot g(x+h)}{g(x+h) \cdot g(x) \cdot h}}$$

Now, we can add and subtract the product of f(x) and g(x), which is $latex f(x)g(x)$, to the numerator $latex f(x+h) \cdot g(x) – f(x) \cdot g(x+h)}{g(x+h)$. Hence, we have

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \lim \limits_{h \to 0} {\frac{f(x+h) \cdot g(x) + f(x) \cdot g(x) – f(x) \cdot g(x) – f(x) \cdot g(x+h)}{g(x+h) \cdot g(x) \cdot h}}$$

Given that $latex + f(x) \cdot g(x) – f(x) \cdot g(x) = 0$, we didn’t change the equation at all.

Re-arranging the previous equation, we have

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \lim \limits_{h \to 0} {\frac{f(x+h) \cdot g(x) – f(x) \cdot g(x) – f(x) \cdot g(x+h) + f(x) \cdot g(x)}{g(x+h) \cdot g(x) \cdot h}}$$

Now, we can further simplify the previous equation by factoring the numerator:

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \lim \limits_{h \to 0} {\frac{g(x) \cdot (f(x+h) – f(x)) – f(x) \cdot (g(x+h) – g(x))}{g(x+h) \cdot g(x) \cdot h}}$$

Then we can further re-arrange the equation like this:

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \lim \limits_{h \to 0} {\left(\frac{1}{g(x+h) \cdot g(x)}\right) \hspace{1.15 pt} \cdot \hspace{1.15 pt} \bigg[\left(g(x) \cdot \left(\frac{f(x+h) – f(x)}{h}\right)\right) – \hspace{1.15 pt} \left(f(x) \cdot \left(\frac{g(x+h) – g(x)}{h}\right)\right)\bigg]}$$

so that we can algebraically manipulate it in a way necessary to prove the quotient rule.

By applying the properties of limits to solve the equation, we have

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \lim \limits_{h \to 0} {\frac{1}{g(x+h) \cdot g(x)}} \hspace{1.15 pt} \cdot \hspace{1.15 pt} \bigg[ \lim \limits_{h \to 0} {g(x)} \hspace{1.15 pt} \cdot \hspace{1.15 pt} \lim \limits_{h \to 0} {\frac{f(x+h) – f(x)}{h}} – \hspace{1.15 pt} \lim \limits_{h \to 0} {f(x)} \hspace{1.15 pt} \cdot \hspace{1.15 pt} \lim \limits_{h \to 0} {\frac{g(x+h) – g(x)}{h}} \bigg]$$

Then, we can solve the limits by recognizing that the first part of each term is simply equal to the functions $latex g(x)$ and $latex f(x)$ respectively and the second part of each term is the limit derivative of $latex f(x)$ and $latex g(x)$ respectively. Therefore, we have:

$$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \left(\frac{1}{g(x) \cdot g(x)}\right) \hspace{1.15 pt} \cdot \hspace{1.15 pt} \Big[ \left( g(x) \cdot \frac{d}{dx}(f(x)) \right) – \hspace{1.15 pt} \left(f(x)\cdot \frac{d}{dx}(g(x)) \right) \Big]$$

By simplifying algebraically, we have

$$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x) \cdot \frac{d}{dx}(f(x)) \hspace{1.15 pt} – \hspace{1.15 pt} f(x) \cdot \frac{d}{dx}(g(x))}{(g(x))^2}$$

or it can be simply illustrated as

$$\left(\frac{f}{g}\right)'(x) = \frac{g(x) \hspace{1.15 pt} \cdot \hspace{1.15 pt} f'(x) \hspace{2.3 pt} – \hspace{2.3 pt} f(x) \hspace{1.15 pt} \cdot \hspace{1.15 pt} g'(x)}{(g(x))^2}$$

which is now The Quotient Rule Formula.

## Proof of The Quotient Rule Using The Product andChain Rules

Another way that might make the quotient rule easier to prove and formulate is by applying the product and chain rules’ formulas. Hence, you are highly recommended to be familiarized with the topics, The Chain Rule Formula and The Product Formula as a pre-requisite to better understand this proof.

We can recall that the product rule formula is

$$(fg)'(x) = f(x) \cdot g'(x) + g(x) \cdot f'(x)$$

In addition to the product rule, we also recall that the chain rule formula is

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

Now, if we are given two functions f(x) and g(x) and then, we are asked to get the derivative of $latex \frac{f}{g}(x)$; we have

$$\left(\frac{f}{g}\right)’ (x) = \frac{d}{dx} \left(\frac{f(x)}{g(x)}\right)$$

By re-writing the denominator of the fraction into exponential form, we have

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{d}{dx} (f(x) \cdot (g(x))^{-1})$$

Now, we can derive the right hand side of the equation by applying the product rule formula:

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = f(x) \cdot \frac{d}{dx} (g(x))^{-1}) + (g(x))^{-1} \cdot \frac{d}{dx}(f(x))$$

To derive $\frac{d}{dx} (g(x))^{-1}$, we need to use the chain rule formula. Hence we have

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = f(x) \cdot \left[(-1) \cdot (g(x))^{-2} \cdot \frac{d}{dx}(g(x)) \right] + (g(x))^{-1} \cdot \frac{d}{dx}(f(x))$$

By applying all applicable operations, getting the least common denominator, re-writing the negative exponent into fractional form, and following the rules of fractions, we have

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = f(x) \cdot \left[(-1) \cdot (g(x))^{-2} \cdot (g'(x)) \right] + (g(x))^{-1} \cdot (f'(x))$$

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = -f(x) \cdot (g(x))^{-2} \cdot (g'(x)) + (g(x))^{-1} \cdot (f'(x))$$

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = -f(x) \cdot \frac{1}{(g(x))^2} \cdot (g'(x)) + \frac{1}{g(x)} \cdot (f'(x))$$

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{-f(x) \cdot (g'(x))}{(g(x))^2} + \frac{(f'(x))}{g(x)}$$

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = -\frac{f(x) \cdot (g'(x))}{(g(x))^2} + \frac{g(x) \cdot (f'(x))}{(g(x))^2}$$

$$\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{g(x) \cdot (f'(x))}{(g(x))^2} \hspace{1.15 pt} – \hspace{1.15 pt} \frac{f(x) \cdot (g'(x))}{(g(x))^2}$$

Then finally, we get

$$\left(\frac{f}{g}\right)'(x) = \frac{g(x) \hspace{1.15 pt} \cdot \hspace{1.15 pt} f'(x) \hspace{2.3 pt} – \hspace{2.3 pt} f(x) \hspace{1.15 pt} \cdot \hspace{1.15 pt} g'(x)}{(g(x))^2}$$

which is now The Quotient Rule Formula.

## Proof of The Quotient Rule Using Implicit Differentiation

This is actually the shortest method of proving the quotient rule formula considering you are familiarized with the topics, the Product Rule, and Implicit Differentiation.

We can recall that implicit differentiation is used for functions in a more complicated form in which it is difficult or impossible to express f(x) or y explicitly in terms of x. For instance, we are given an equation:

$latex y = \frac{u}{v}$

and then we are asked to derive $latex y$. By deriving $latex y$, we have

$latex y’ = \left(\frac{u}{v}\right)’$

But assuming we cannot further simplify our equation algebraically and still do not know the quotient rule formula, we can do derive it by applying implicit differentiation.

To implicitly differentiate our given equation, we must first algebraically cross multiply the denominator of the right-hand side to the left-hand side of the equation. Hence, we have

$latex vy = u$

Then, we derive the whole equation in terms of the variable $latex x$:

$latex \frac{d}{dx} (vy) = \frac{d}{dx} (u)$

To derive the left-hand side of the equation, we will use the product rule. We will also treat $latex v$ and $latex y$ as variables and not constants. By doing so, we have

$latex vy’ + yv’ = \frac{d}{dx} (u)$

How about $latex u$? How do we derive $latex u$ in terms of variable $latex x$ in this case? Just like $latex v$ and $latex y$, we will treat $latex u$ as a variable, not as a constant. By doing so, we have

$latex vy’ + yv’ = u’$

Since we are asked to get $latex y’$, we need to equate our equation in terms of $latex y’$. By doing so, we have

$latex y’ = \frac{u’-yv’}{v}$

But what is $latex y$? We can recall from the beginning that $latex y = \frac{u}{v}$. By substituting $latex y$ to our derived equation, we have

$latex y’ = \frac{u’-\left(\frac{u}{v}\right) \cdot v’}{v}$

By simplifying algebraically, getting the least common denominator, and applying the rules of fractions, we have

$latex y’ = \frac{u’-\left(\frac{u}{v}\right) \cdot v’}{v}$

$latex y’ = \frac{u’-\left(\frac{uv’}{v}\right)}{v}$

$latex y’ = \frac{\left(\frac{vu’}{v}\right)-\left(\frac{uv’}{v}\right)}{v}$

$latex \frac{d}{dx}(\frac{u}{v}) = \frac{vu’ \hspace{2.3 pt} – \hspace{2.3 pt} uv’}{(v) \cdot (v)}$

Then finally, we have

which is now The Quotient Rule Formula.