17Calculus Derivatives - Quotient Rule
When you are first learning the quotient rule, it is a good idea to write out intermediate steps. This rule is easy to get confused about, so writing out intermediate steps will help you get your head around the details. Later, you will be able to do more in your head and less on paper.
How To Use The Quotient Rule |
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The quotient rule looks like this. If you have a function \(\displaystyle{f(x) = \frac{n(x)}{d(x)}} \), the quotient rule says the derivative is \(\displaystyle{ f'(x) = \frac{d \cdot n' - n \cdot d'}{d^2} }\). In this equation, we have used \(n(x)\) to denote the expression in the numerator and \(d(x)\) to denote the expression in the denominator.
Okay, so this seems pretty straight-forward. Let's do an example. Try this on your own before looking at the solution.
Calculate the derivative of \( \displaystyle{f(x) = \frac{x+5}{x^2}} \).
\(\displaystyle{ f'(x) = \frac{-(x+10)}{x^3}}\)
Problem Statement - Calculate the derivative of \( \displaystyle{f(x) = \frac{x+5}{x^2}} \).
Solution - In this case \( n(x) = x+5 \) and \( d(x) = x^2 \).
\(\displaystyle{ f'(x) = \frac{(x^2)(1) - (x+5)(2x)}{(x^2)^2} = }\)
\(\displaystyle{ \frac{x[x-2(x+5)]}{x^4} = }\)
\(\displaystyle{ \frac{x-2x-10}{x^3} = }\)
\(\displaystyle{ \frac{-(x+10)}{x^3}}\)
Final Answer |
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\(\displaystyle{ f'(x) = \frac{-(x+10)}{x^3}}\) |
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A Note About Simplifying | |
Things To Watch For |
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There are a few things we see students do quite often that we want to warn you about.
First, go back to the example and notice that we take the denominator and assign it directly to d(x). I have seen students think \( d(x) = x^{-2} \). This is NOT the case. We do not take the denominator, move it to the numerator and call that d(x). So be very careful here.
Second, again, go back to the example and notice that, when simplifying, the first thing we do in the numerator is look for a common factor between the two terms. In the example, we had a factor of x. We factored it out and canceled it with an x in the denominator. This happens quite often when using the quotient rule.
Third, the quotient rule itself is not that difficult to do. The thing that will probably trip you up the most is the algebra you have to do with simplifying. So this technique will often challenge you to remember and use your algebra rules related to factoring and powers.
Before working some practice problems, take a few minutes and watch this video showing a proof of the quotient rule.
MIP4U - Proof of the Quotient Rule [6min-27secs]
video by MIP4U |
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Okay, it is time for some practice problems. After that, the chain rule is next. |
Practice
Instructions - - Unless otherwise instructed, calculate the derivatives of the following functions using the quotient rule, giving your answers in simplified form.
Here are some problems that use only the quotient rule and the basic rules discussed on the main derivatives page (power rule, constant rule and constant multiple rule).
Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{f(x)=\frac{x+1}{x-1}}\)
Solution |
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949 video
video by Krista King Math |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{f(x)=\frac{x^2-4}{x^2+4}}\)
Solution |
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950 video
video by Krista King Math |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{f(x)=\frac{3}{x^2+x+1}}\)
Solution |
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951 video
video by Krista King Math |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{f(x)=\frac{1}{x+1}-\frac{1}{x-1}}\)
Solution |
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948 video
video by Krista King Math |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{\frac{x^3-5x^2+7x+3}{x^2+9}}\).
Final Answer |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{\frac{x^3-5x^2+7x+3}{x^2+9}}\).
Solution |
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\(\displaystyle{\frac{d}{dx}\left[\frac{x^3-5x^2+7x+3}{x^2+9}\right] }\) |
Apply the quotient rule. |
\(\displaystyle{\frac{(x^2+9)d[x^3-5x^2+7x+3]/dx - (x^3-5x^2+7x+3)d[x^2+9]/dx}{(x^2+9)^2} }\) |
Take the derivatives. |
\(\displaystyle{\frac{(x^2+9)(3x^2-10x+7)-(x^3-5x^2+7x+3)(2x)}{(x^2+9)^2} }\) |
Look for common terms in the numerator. In this case, there are not any, so just multiply out. |
\(\displaystyle{\frac{(3x^4 -10x^3+7x^2+27x^2-90x+63)-(2x^4-10x^3+14x^2+6x)}{(x^2+9)^2} }\) |
\(\displaystyle{\frac{x^4+20x^2-96x+63}{(x^2+9)^2}}\) |
After completing all derivatives, we always check for common terms between the two main factors in the numerator to factor out. In this case, there were none, so our only option was to multiply out. Notice we leave the denominator in its most compact and factored form.
Final Answer |
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\( \displaystyle{ \frac{d}{dx} \left[ \frac{x^3-5x^2+7x+3}{x^2+9} \right] = \frac{x^4+20x^2-96x+63}{(x^2+9)^2} } \) |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{\frac{x^2-3x+7}{\sqrt{x}}}\).
Final Answer |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{\frac{x^2-3x+7}{\sqrt{x}}}\).
Solution |
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\(\begin{array}{rcl}
& & \frac{d}{dx} \left[ \frac{x^2-3x+7}{\sqrt{x}} \right ] \\
& = & \frac{x^{1/2}(2x-3) - (x^2-3x+7)(1/2)x^{-1/2}}{x} \\
& = & \frac{x(2x-3) - (1/2)(x^2-3x+7)}{x^{3/2}} \\
& = & \frac{2x(2x-3)-(x^2-3x+7)}{2x^{3/2}} \\
& = & \frac{3x^2-3x-7}{2x^{3/2}}
\end{array}\)
In the first line, we used the quotient rule. After that, just algebra. The tricky algebra here involves \(x^{-1/2}\). In the second line, we have x's in both of the main terms in the numerator. We need to factor those out. So we multiply the numerator and the denominator by \( x^{1/2} \). We chose \( x^{1/2} \) since our goal is to have all positive powers in our answer and multiplying \( x^{-1/2} \) by \( x^{1/2} \) gives us \( x^{-1/2} \cdot x^{1/2} = x^{-1/2+1/2} = x^0 = 1 \). This gets rid of the \( x^{-1/2} \) in the second term.
Final Answer |
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\(\displaystyle{\frac{3x^2-3x-7}{2x^{3/2}}}\) |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{y=\frac{t^2+2}{t^4-3t^2+1}}\)
Solution |
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946 video
video by Krista King Math |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{f(x)=\frac{x-1}{x^2+2x+1}}\)
Solution |
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947 video
video by Krista King Math |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{y=\frac{45}{5+x+\sqrt{x}}}\)
Solution |
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952 video
video by Krista King Math |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{y=\frac{x^2+1}{x^5+x}}\)
Solution |
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954 video
video by PatrickJMT |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{f(x)=\frac{x^2+3x}{x+4}}\)
Solution |
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1313 video
video by PatrickJMT |
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Here are a couple of problems that are a little more difficult but, after working the above problems successfully, you should have no problem with these.
Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{f(x)=\frac{x}{x+c/x}}\)
Solution |
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953 video
video by Krista King Math |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{ h(x)=\frac{2x^3~k(x)}{3x+2} }\)
Hint |
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The function \(k(x)\) is not known. So when you take the derivative just write \(k'(x)\).
Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{ h(x)=\frac{2x^3~k(x)}{3x+2} }\)
Final Answer |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{ h(x)=\frac{2x^3~k(x)}{3x+2} }\)
Hint |
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The function \(k(x)\) is not known. So when you take the derivative just write \(k'(x)\).
Solution |
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2225 video
video by MathTV |
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Final Answer |
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\(\displaystyle{ h'(x) = \frac{2x^2[ 6k(x)(x+1) + xk'(x)(3x+2) ]}{(3x+2)^2} }\) |
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These problems require you to know how to take the derivative of exponential functions.
Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{\frac{7e^t}{5-9e^t}}\).
Final Answer |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{\frac{7e^t}{5-9e^t}}\).
Solution |
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\(
\begin{array}{rcl}
& & \frac{d}{dt} \left[ \frac{7e^t}{5-9e^t} \right] \\
& = & \frac{(5-9e^t)d[7e^t]/dt - (7e^t)d[5-9e^t]/dt}{(5-9e^t)^2} \\
& = & \frac{(5-9e^t)(7e^t) - 7e^t(-9e^t)}{(5-9e^t)^2} \\
& = & \frac{(7e^t)(5-9e^t +9e^t)}{(5-9e^t)^2} \\
& = & \frac{35e^t}{(5-9e^t)^2} \end{array} \)
In the first step, we used the quotient rule. Following that, we used algebra to factor and simplify. Notice that between lines 3 and 4, we factored out \( 7e^t \) before multiplying out the numerator.
Final Answer |
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\( \displaystyle{ \frac{35e^t}{(5-9e^t)^2} }\) |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{f(x)=\frac{7x^2+1}{e^x}}\)
Solution |
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945 video
video by Krista King Math |
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These problems require you to know how to take the derivative of trig functions.
Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{\frac{\sin(x)}{1+\cos(x)}}\).
Final Answer |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{\frac{\sin(x)}{1+\cos(x)}}\).
Solution |
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\(\displaystyle{ \frac{d}{dx}\left[ \frac{\sin(x)}{1+\cos(x)} \right] }\) |
\(\displaystyle{ \frac{[1+\cos(x)]d[\sin(x)]/dx - [\sin(x)]d[1+\cos(x)]/dx}{[1+\cos(x)]^2} }\) |
\(\displaystyle{ \frac{[1+\cos(x)][\cos(x)] - [\sin(x)][-\sin(x)]}{[1+\cos(x)]^2} }\) |
\(\displaystyle{ \frac{\cos(x)+\cos^2(x)+\sin^2(x)}{[1+\cos(x)]^2} }\) |
Use the identity \( \sin^2(x) + \cos^2(x) = 1 \). |
\(\displaystyle{ \frac{1+\cos(x)}{[1+\cos(x)]^2} }\) |
\(\displaystyle{ \frac{1}{1+\cos(x)} }\) |
Final Answer |
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\(\displaystyle{ \frac{d}{dx}\left[ \frac{\sin(x)}{1+\cos(x)} \right] = \frac{1}{1+\cos(x)}}\) |
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Problem Statement |
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Use the quotient rule to calculate the derivative of \(\displaystyle{y=\frac{\tan(x)}{x^{3/2}+5x}}\)
Solution |
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955 video
video by PatrickJMT |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
Some of the practice problems require you to know the following rules also (in their basic form, not including the chain rule). If you don't know one or more of these rules, no worries. You can skip those problems and come back to them later. |
You do NOT need to know the chain rule for anything on this page, including practice problems. For practice problems using the quotient rule and chain rule, see the chain rule page. |
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