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2.9 The Quotient Rule

3 min readjune 18, 2024

2.9 The Quotient Rule

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!

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Image Courtesy of OnLine Math Learning