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2.9 The Quotient Rule

3 min readfebruary 15, 2024

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As the complexity of the functions you encounter increases, applying the basic Power Rule correctly becomes more difficult. Like the Product Rule explored in the previous lesson, the Quotient Rule simplifies the process of differentiating the quotient of functions, allowing for the differentiation of larger and more complex functions. 📈


⛳ Quotient Rule Definition

The Quotient Rule is as stated below:

ddx[f(x)g(x)]=g(x)ddxf(x)f(x)ddxg(x)(g(x))2\frac{d}{dx} \Bigg[\frac{f(x)}{g(x)}\Bigg] = \frac{g(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}g(x)}{(g(x))^2}

Similar to the Product Rule, the Quotient Rule uses the original functions and their derivatives to compute the final derivative. Rather than adding the products of the functions and their derivatives, they are subtracted and divided by the square of the denominator function (in this case, g(x)g(x)). If this sounds and looks a little too complicated, you can substitute variables to represent the functions as shown below.

Given that u=f(x)\textcolor{red}{u = f(x)} and v=g(x)\textcolor{green}{v = g(x)}, then the Quotient Rule can also be written as:

ddx[f(x)g(x)]=ddx(uv)=vuuvv2\frac{d}{dx} \Bigg[\frac{\textcolor{red}{f(x)}}{\textcolor{green}{g(x)}}\Bigg] = \frac{d}{dx} (\frac{\textcolor{red}{u}}{\textcolor{green}{v}}) = \frac{\textcolor{green}{v}\textcolor{purple}{u'}-\textcolor{red}{u}\textcolor{blue}{v'}}{\textcolor{green}{v^2}}

Note: The Quotient Rule can only be used when the equation is in the form of f(x)g(x)\frac{f(x)}{g(x)} AND both functions are differentiable. ✅

✏️ Quotient Rule Walkthrough

Now that we know what the Quotient Rule is and how to use it, let's apply it to an example! 📚

🔢 Suppose you are given the rational function shown below:

y=x2+x2x3+6y=\frac{x^2+x-2}{x^3+6}

First, notice how it is already in the form of f(x)g(x)\frac{f(x)}{g(x)}, so we can apply the Quotient Rule to calculate its derivative. If we say that f(x)=x2+x2f(x) = x^2+x-2 and g(x)=x3+6g(x) = x^3 + 6, then:

dydx[x2+x2x3+6]=(x3+6)ddx(x2+x2)(x2+x2)ddx(x3+6)(x3+6)2\frac{dy}{dx} \Bigg[\frac{x^2+x-2}{x^3+6}\Bigg] = \frac{(x^3+6)*\frac{d}{dx}(x^2+x-2)-(x^2+x-2)*\frac{d}{dx}(x^3+6)}{(x^3+6)^2}

Using the formula given by the Quotient Rule, the derivative of this rational function is rewritten, and can be simplified as follows: ✍️

dydx[x2+x2x3+6]=(x3+6)(2x+1)(3x2)(x2+x2)(x3+6)2\frac{dy}{dx} \Bigg[\frac{x^2+x-2}{x^3+6}\Bigg] = \frac{(x^3+6)(2x+1)-(3x^2)(x^2+x-2)}{(x^3+6)^2}

Here, we rewrote the rational function with the derivatives of the numerator and denominator. After this, you can use algebra to combine and further simplify the derivative:

dydx[x2+x2x3+6]=(2x4+x3+12x+6)(3x4+3x36x2)(x3+6)2\frac{dy}{dx} \Bigg[\frac{x^2+x-2}{x^3+6}\Bigg] = \frac{(2x^4+x^3+12x+6)-(3x^4+3x^3-6x^2)}{(x^3+6)^2}

By multiplying binomial and combining like terms, our final answer is:

dydx[x2+x2x3+6]=(x42x3+6x2+12x+6)(x3+6)2\frac{dy}{dx} \Bigg[\frac{x^2+x-2}{x^3+6}\Bigg] =\frac{(-x^4-2x^3+6x^2+12x+6)}{(x^3+6)^2}

From here, the derivative can be simplified further, but this is only sometimes necessary. Do try to simplify the derivative if possible. 💡

This rule can also be applied to other types of functions besides algebraic functions, such as trigonometric functions (sin(x),cos(x),tan(x)\sin(x), \cos(x), \tan(x) , etc) and exponential functions (ex)(e^x). Try the next two examples of applying the Quotient Rule to differentiate these types of functions! 🏁


🧮 Quotient Rule: Practice Problems

Give the following problems a try yourself before you see the way we walk through them!

Quotient Rule: Example 1

Suppose you are given the rational function below:

y=ex1+x2y = \frac{e^x}{1+x^2}

In applying the Quotient Rule, we see that f(x)=exf(x) = e^x and g(x)=1+x2g(x) = 1+x^2. Recall that the derivative of exe^x is just exe^x.

dydx[ex1+x2]=(1+x2)ddx(ex)(ex)ddx(1+x2)(1+x2)2\frac{dy}{dx} \Bigg[\frac{e^x}{1+x^2}\Bigg] = \frac{(1+x^2)*\frac{d}{dx}(e^x)-(e^x)*\frac{d}{dx}(1+x^2)}{(1+x^2)^2}

Therefore,

dydx[ex1+x2]=(1+x2)(ex)(ex)(2x)(1+x2)2\frac{dy}{dx} \Bigg[\frac{e^x}{1+x^2}\Bigg] = \frac{(1+x^2)*(e^x)-(e^x)*(2x)}{(1+x^2)^2}

This can either be left as is, but can also be simplified for better clarity:

dydx[ex1+x2]=ex(12x+x2)(1+x2)2=ex(1x)2(1+x2)2\frac{dy}{dx} \Bigg[\frac{e^x}{1+x^2}\Bigg] = \frac{e^x(1-2x+x^2)}{(1+x^2)^2} = \frac{e^x(1-x)^2}{(1+x^2)^2}

This makes the final answer to be ex(1x)2(1+x2)2\frac{e^x(1-x)^2}{(1+x^2)^2}! Now let’s try using the Quotient Rule with trigonometric functions. ⬇️

Quotient Rule: Example 2

Suppose you are given the following rational function below:

y=sin(x)1+cos(x)y = \frac{sin(x)}{1+cos(x)}

Again, as we apply the Quotient Rule, we can say that f(x)=sin(x)f(x) = sin(x) and g(x)=1+cos(x)g(x) = 1 + cos(x). For this specific problem, you would need to recall that…

ddx(sin(x))=cos(x)\frac{d}{dx}(sin(x)) = cos(x)
ddx(cos(x))=sin(x)\frac{d}{dx}(cos(x)) = -sin(x)

For similar problems like this, it would be best to memorize the derivatives of all the trigonometric functions. Check out this guide if you need a review!

In doing so, we can rewrite this as:

dydx[sin(x)1+cos(x)]=1+cos(x)ddx(sin(x))+sin(x)ddx(1+cos(x))(1+cos(x))2\frac{dy}{dx}\Bigg[\frac{sin(x)}{1+cos(x)}\Bigg] = \frac{1+cos(x)*\frac{d}{dx}(sin(x)) + sin(x)*\frac{d}{dx}(1+cos(x))}{(1+cos(x))^2}

Therefore,

dydx[sin(x)1+cos(x)]=sin(x)(1+cos(x))+sin(x)(sin(x))(1+cos(x))2\frac{dy}{dx}\Bigg[\frac{sin(x)}{1+cos(x)}\Bigg] = \frac{sin(x)(1+cos(x)) + sin(x)(-sin(x))}{(1+cos(x))^2}

After simplifying, the final answer would be:

dydx[sin(x)1+cos(x)]=sin(x)(1+cos(x)sin(x))(1+cos(x))2\frac{dy}{dx}\Bigg[\frac{sin(x)}{1+cos(x)}\Bigg] = \frac{sin(x)(1+cos(x)-sin(x))}{(1+cos(x))^2}

💫 Closing

Great work! Using the Quotient Rule, finding the derivative of rational functions becomes a lot easier. 🎉

Here’s a quick summary for you!

Untitled

Image Courtesy of OnLine Math Learning

Key Terms to Review (4)

Chain Rule

: The chain rule is a formula used to find the derivative of a composition of two or more functions. It states that the derivative of a composite function is equal to the derivative of the outermost function times the derivative of the innermost function.

Denominator

: The denominator is the bottom part of a fraction that represents the total number of equal parts into which a whole is divided.

Function

: A relationship between two sets where each input (domain) value corresponds to exactly one output (range) value.

Quotient Rule

: The quotient rule is a formula used to find the derivative of a quotient or division between two functions. It states that the derivative of a fraction is equal to (the denominator times the derivative of numerator) minus (the numerator times the derivative of denominator), all divided by (the square root of denominator squared).

2.9 The Quotient Rule

3 min readfebruary 15, 2024

Attend a live cram event

Review all units live with expert teachers & students

As the complexity of the functions you encounter increases, applying the basic Power Rule correctly becomes more difficult. Like the Product Rule explored in the previous lesson, the Quotient Rule simplifies the process of differentiating the quotient of functions, allowing for the differentiation of larger and more complex functions. 📈


⛳ Quotient Rule Definition

The Quotient Rule is as stated below:

ddx[f(x)g(x)]=g(x)ddxf(x)f(x)ddxg(x)(g(x))2\frac{d}{dx} \Bigg[\frac{f(x)}{g(x)}\Bigg] = \frac{g(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}g(x)}{(g(x))^2}

Similar to the Product Rule, the Quotient Rule uses the original functions and their derivatives to compute the final derivative. Rather than adding the products of the functions and their derivatives, they are subtracted and divided by the square of the denominator function (in this case, g(x)g(x)). If this sounds and looks a little too complicated, you can substitute variables to represent the functions as shown below.

Given that u=f(x)\textcolor{red}{u = f(x)} and v=g(x)\textcolor{green}{v = g(x)}, then the Quotient Rule can also be written as:

ddx[f(x)g(x)]=ddx(uv)=vuuvv2\frac{d}{dx} \Bigg[\frac{\textcolor{red}{f(x)}}{\textcolor{green}{g(x)}}\Bigg] = \frac{d}{dx} (\frac{\textcolor{red}{u}}{\textcolor{green}{v}}) = \frac{\textcolor{green}{v}\textcolor{purple}{u'}-\textcolor{red}{u}\textcolor{blue}{v'}}{\textcolor{green}{v^2}}

Note: The Quotient Rule can only be used when the equation is in the form of f(x)g(x)\frac{f(x)}{g(x)} AND both functions are differentiable. ✅

✏️ Quotient Rule Walkthrough

Now that we know what the Quotient Rule is and how to use it, let's apply it to an example! 📚

🔢 Suppose you are given the rational function shown below:

y=x2+x2x3+6y=\frac{x^2+x-2}{x^3+6}

First, notice how it is already in the form of f(x)g(x)\frac{f(x)}{g(x)}, so we can apply the Quotient Rule to calculate its derivative. If we say that f(x)=x2+x2f(x) = x^2+x-2 and g(x)=x3+6g(x) = x^3 + 6, then:

dydx[x2+x2x3+6]=(x3+6)ddx(x2+x2)(x2+x2)ddx(x3+6)(x3+6)2\frac{dy}{dx} \Bigg[\frac{x^2+x-2}{x^3+6}\Bigg] = \frac{(x^3+6)*\frac{d}{dx}(x^2+x-2)-(x^2+x-2)*\frac{d}{dx}(x^3+6)}{(x^3+6)^2}

Using the formula given by the Quotient Rule, the derivative of this rational function is rewritten, and can be simplified as follows: ✍️

dydx[x2+x2x3+6]=(x3+6)(2x+1)(3x2)(x2+x2)(x3+6)2\frac{dy}{dx} \Bigg[\frac{x^2+x-2}{x^3+6}\Bigg] = \frac{(x^3+6)(2x+1)-(3x^2)(x^2+x-2)}{(x^3+6)^2}

Here, we rewrote the rational function with the derivatives of the numerator and denominator. After this, you can use algebra to combine and further simplify the derivative:

dydx[x2+x2x3+6]=(2x4+x3+12x+6)(3x4+3x36x2)(x3+6)2\frac{dy}{dx} \Bigg[\frac{x^2+x-2}{x^3+6}\Bigg] = \frac{(2x^4+x^3+12x+6)-(3x^4+3x^3-6x^2)}{(x^3+6)^2}

By multiplying binomial and combining like terms, our final answer is:

dydx[x2+x2x3+6]=(x42x3+6x2+12x+6)(x3+6)2\frac{dy}{dx} \Bigg[\frac{x^2+x-2}{x^3+6}\Bigg] =\frac{(-x^4-2x^3+6x^2+12x+6)}{(x^3+6)^2}

From here, the derivative can be simplified further, but this is only sometimes necessary. Do try to simplify the derivative if possible. 💡

This rule can also be applied to other types of functions besides algebraic functions, such as trigonometric functions (sin(x),cos(x),tan(x)\sin(x), \cos(x), \tan(x) , etc) and exponential functions (ex)(e^x). Try the next two examples of applying the Quotient Rule to differentiate these types of functions! 🏁


🧮 Quotient Rule: Practice Problems

Give the following problems a try yourself before you see the way we walk through them!

Quotient Rule: Example 1

Suppose you are given the rational function below:

y=ex1+x2y = \frac{e^x}{1+x^2}

In applying the Quotient Rule, we see that f(x)=exf(x) = e^x and g(x)=1+x2g(x) = 1+x^2. Recall that the derivative of exe^x is just exe^x.

dydx[ex1+x2]=(1+x2)ddx(ex)(ex)ddx(1+x2)(1+x2)2\frac{dy}{dx} \Bigg[\frac{e^x}{1+x^2}\Bigg] = \frac{(1+x^2)*\frac{d}{dx}(e^x)-(e^x)*\frac{d}{dx}(1+x^2)}{(1+x^2)^2}

Therefore,

dydx[ex1+x2]=(1+x2)(ex)(ex)(2x)(1+x2)2\frac{dy}{dx} \Bigg[\frac{e^x}{1+x^2}\Bigg] = \frac{(1+x^2)*(e^x)-(e^x)*(2x)}{(1+x^2)^2}

This can either be left as is, but can also be simplified for better clarity:

dydx[ex1+x2]=ex(12x+x2)(1+x2)2=ex(1x)2(1+x2)2\frac{dy}{dx} \Bigg[\frac{e^x}{1+x^2}\Bigg] = \frac{e^x(1-2x+x^2)}{(1+x^2)^2} = \frac{e^x(1-x)^2}{(1+x^2)^2}

This makes the final answer to be ex(1x)2(1+x2)2\frac{e^x(1-x)^2}{(1+x^2)^2}! Now let’s try using the Quotient Rule with trigonometric functions. ⬇️

Quotient Rule: Example 2

Suppose you are given the following rational function below:

y=sin(x)1+cos(x)y = \frac{sin(x)}{1+cos(x)}

Again, as we apply the Quotient Rule, we can say that f(x)=sin(x)f(x) = sin(x) and g(x)=1+cos(x)g(x) = 1 + cos(x). For this specific problem, you would need to recall that…

ddx(sin(x))=cos(x)\frac{d}{dx}(sin(x)) = cos(x)
ddx(cos(x))=sin(x)\frac{d}{dx}(cos(x)) = -sin(x)

For similar problems like this, it would be best to memorize the derivatives of all the trigonometric functions. Check out this guide if you need a review!

In doing so, we can rewrite this as:

dydx[sin(x)1+cos(x)]=1+cos(x)ddx(sin(x))+sin(x)ddx(1+cos(x))(1+cos(x))2\frac{dy}{dx}\Bigg[\frac{sin(x)}{1+cos(x)}\Bigg] = \frac{1+cos(x)*\frac{d}{dx}(sin(x)) + sin(x)*\frac{d}{dx}(1+cos(x))}{(1+cos(x))^2}

Therefore,

dydx[sin(x)1+cos(x)]=sin(x)(1+cos(x))+sin(x)(sin(x))(1+cos(x))2\frac{dy}{dx}\Bigg[\frac{sin(x)}{1+cos(x)}\Bigg] = \frac{sin(x)(1+cos(x)) + sin(x)(-sin(x))}{(1+cos(x))^2}

After simplifying, the final answer would be:

dydx[sin(x)1+cos(x)]=sin(x)(1+cos(x)sin(x))(1+cos(x))2\frac{dy}{dx}\Bigg[\frac{sin(x)}{1+cos(x)}\Bigg] = \frac{sin(x)(1+cos(x)-sin(x))}{(1+cos(x))^2}

💫 Closing

Great work! Using the Quotient Rule, finding the derivative of rational functions becomes a lot easier. 🎉

Here’s a quick summary for you!

Untitled

Image Courtesy of OnLine Math Learning

Key Terms to Review (4)

Chain Rule

: The chain rule is a formula used to find the derivative of a composition of two or more functions. It states that the derivative of a composite function is equal to the derivative of the outermost function times the derivative of the innermost function.

Denominator

: The denominator is the bottom part of a fraction that represents the total number of equal parts into which a whole is divided.

Function

: A relationship between two sets where each input (domain) value corresponds to exactly one output (range) value.

Quotient Rule

: The quotient rule is a formula used to find the derivative of a quotient or division between two functions. It states that the derivative of a fraction is equal to (the denominator times the derivative of numerator) minus (the numerator times the derivative of denominator), all divided by (the square root of denominator squared).


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© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.