In my studybook from college I've found this problem: differentiate ""ln((1+e^x)/(1+e^(-x)))"" and I don't seem to be able to crack it, how kan I solve this one?

There's even a quicker way! We can first simplify the expression inside the logarithm, by factoring out e^x in the numerator: (1+e^x)/(1+e^(-x)) = (e^x)(e^(-x)+1)/(1+e^(-x)) = (e^x)(1+e^(-x))/(1+e^(-x)) = e^x. Therefore, we have d/dx of (ln((1+e^x)/(1+e^(-x)))) = d/dx of (ln(e^x)) = d/dx of (x) = 1. Amazing!

In the first Checking Your Understanding above, isn't the d/dx of e^x = xe^(x-1) ? Else what is the rule applied to e^x?

The power rule only applies when the exponent is a constant. For example x³ or x ⁻². It does not apply when the exponent is a variable. For example 2ˣ or, as in this case eˣ. There's quite a long section on the derivatives of exponential functions, here's the video on eˣ https://www.khanacademy.org/math/ap-calculus-ab/ab-derivatives-advanced/ab-diff-exp/v/derivative-of-ex

What would be the reason that one couldn't simply take a quotient (for example (x^2 - 3)/x^4 and simply make it the product (x^2-3)*x^-4 and then apply the product rule? Of course, they have different answers because the product rule does not factor in the square of the denominator, but what rule would be broken by doing so?

Yes, you can express (x^2 - 3)/x^4 as the product (x^2 - 3) * x^-4 and use the product rule to take the derivative. No rule is broken here. Your answer might not appear the same as if you used the quotient rule to differentiate (x^2 - 3)/x^4, but it should end up mathematically equivalent. If not, then you've made an algebraic or differentiation error somewhere. In fact, the product and chain rules can be used to *derive* the quotient rule: d/dx of [f(x)/g(x)] = d/dx of {f(x)*[g(x)]^(-1)} = f(x)*(-1)[g(x)]^(-2)*g'(x) + f'(x)*[g(x)]^(-1) = [g(x)]^(-2) * [-f(x)*g'(x) + f'(x)*g(x)] = [g(x)*f'(x) - f(x)*g'(x)]/[g(x)]^2. Have a blessed, wonderful day!

should it not be written like this [d/dx(g(x)) .f(x)- d/d(x)(f(x)).g(x)]/[g(x)]^2, i found this lesson extremely baffling as this opposes what i learned from my textbook, which formula is the right one? any suggestion to improve my understanding,please?

Your textbook is incorrect; from all other sources, it is indeed f'g - g'f. Try deriving it yourself using the product rule and chain rule.

For question 1 I used x^2 as my f(x) instead of e^x and ended up getting [e^x(2-x)]/(x^3). Would this be incorrect?

Division isn't commutative like multiplication, so if you switch the positions of the numbers you're dividing, you'll get a different answer. From this, it follows that the derivative of one function divided by a second one would be different than the derivative of the second divided by the first. You don't have to be careful about this when doing the product rule, but when doing the quotient rule, remember that you subtract term with the derivative of the bottom function, and divide by the bottom function squared. Anything else shouldn't give you the right answer, and (e^x (2-x)) / x^3 would be incorrect.

In problem 2 of example 1 you state that the answer is (−xsin(x)+3cos(x))/x^4 however the actual answer is (which the site recognizes) is flip out that middle "+" for a "-"

The first minus sign in the solution has been brought out the front of the fraction, so they are stating the answer is −(xsin(x)+3cos(x))/x^4, which is equal to (−xsin(x)-3cos(x))/x^4, rather than (−xsin(x)+3cos(x))/x^4

Main content

AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 2

Lesson 10: The quotient rule

Quotient rule review

AP.CALC:

FUN‑3 (EU)

FUN‑3.B (LO)

FUN‑3.B.2 (EK)

Google Classroom

Review your knowledge of the Quotient rule for derivatives, and use it to solve problems.

What is the Quotient rule?

The Quotient rule tells us how to differentiate expressions that are the quotient of two other, more basic, expressions:

start fraction, d, divided by, d, x, end fraction, open bracket, start fraction, f, left parenthesis, x, right parenthesis, divided by, g, left parenthesis, x, right parenthesis, end fraction, close bracket, equals, start fraction, start fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, x, right parenthesis, close bracket, dot, g, left parenthesis, x, right parenthesis, minus, f, left parenthesis, x, right parenthesis, dot, start fraction, d, divided by, d, x, end fraction, open bracket, g, left parenthesis, x, right parenthesis, close bracket, divided by, open bracket, g, left parenthesis, x, right parenthesis, close bracket, squared, end fraction

Basically, you take the derivative of f multiplied by g, subtract f multiplied by the derivative of g, and divide all that by open bracket, g, left parenthesis, x, right parenthesis, close bracket, squared.

Want to learn more about the Quotient rule? Check out this video.

What problems can I solve with the Quotient rule?

Example 1

Consider the following differentiation of start fraction, sine, left parenthesis, x, right parenthesis, divided by, x, squared, end fraction:

\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac{\sin(x)}{x^2}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(\sin(x))x^2-\sin(x)\dfrac{d}{dx}(x^2)}{(x^2)^2}&&\gray{\text{Quotient rule}} \\\\ &=\dfrac{\cos(x)\cdot x^2-\sin(x)\cdot 2x}{(x^2)^2}&&\gray{\text{Differentiate }\sin(x)\text{ and }x^2} \\\\ &=\dfrac{x\left(x\cos(x)-2\sin(x)\right)}{x^4}&&\gray{\text{Simplify}} \\\\ &=\dfrac{x\cos(x)-2\sin(x)}{x^3}&&\gray{\text{Cancel common factors}} \end{aligned}

Check your understanding

Problem 1

Current

f, left parenthesis, x, right parenthesis, equals, start fraction, x, squared, divided by, e, start superscript, x, end superscript, end fraction

f, prime, left parenthesis, x, right parenthesis, equals

Want to try more problems like this? Check out this exercise.

Example 2

Suppose we are given this table of values:

x	f, left parenthesis, x, right parenthesis	g, left parenthesis, x, right parenthesis	f, prime, left parenthesis, x, right parenthesis	g, prime, left parenthesis, x, right parenthesis
4	minus, 4	minus, 2	0	8

H, left parenthesis, x, right parenthesis is defined as start fraction, f, left parenthesis, x, right parenthesis, divided by, g, left parenthesis, x, right parenthesis, end fraction, and we are asked to find H, prime, left parenthesis, 4, right parenthesis.

The Quotient rule tells us that H, prime, left parenthesis, x, right parenthesis is start fraction, f, prime, left parenthesis, x, right parenthesis, g, left parenthesis, x, right parenthesis, minus, f, left parenthesis, x, right parenthesis, g, prime, left parenthesis, x, right parenthesis, divided by, open bracket, g, left parenthesis, x, right parenthesis, close bracket, squared, end fraction. This means H, prime, left parenthesis, 4, right parenthesis is start fraction, f, prime, left parenthesis, 4, right parenthesis, g, left parenthesis, 4, right parenthesis, minus, f, left parenthesis, 4, right parenthesis, g, prime, left parenthesis, 4, right parenthesis, divided by, open bracket, g, left parenthesis, 4, right parenthesis, close bracket, squared, end fraction. Now let's plug the values from the table in the expression:

\begin{aligned} H'(4)&=\dfrac{f'(4)g(4)-f(4)g'(4)}{[g(4)]^2} \\\\ &=\dfrac{(0)(-2)-(-4)(8)}{(-2)^2} \\\\ &=\dfrac{32}{4} \\\\ &=8 \end{aligned}

Check your understanding

Problem 1

Current

x	g, left parenthesis, x, right parenthesis	h, left parenthesis, x, right parenthesis	g, prime, left parenthesis, x, right parenthesis	h, prime, left parenthesis, x, right parenthesis
minus, 2	4	1	minus, 1	2

F, left parenthesis, x, right parenthesis, equals, start fraction, g, left parenthesis, x, right parenthesis, divided by, h, left parenthesis, x, right parenthesis, end fraction

F, prime, left parenthesis, minus, 2, right parenthesis, equals

Want to try more problems like this? Check out this exercise.

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J Kuijpers
Posted 6 years ago. Direct link to J Kuijpers's post “In my studybook from coll...”
In my studybook from college I've found this problem:
differentiate ""ln((1+e^x)/(1+e^(-x)))""
and I don't seem to be able to crack it, how kan I solve this one?
Button navigates to signup pageButton navigates to signup page
(14 votes)
Answer
- Ian Pulizzotto
  Posted 6 years ago. Direct link to Ian Pulizzotto's post “There's even a quicker wa...”
  There's even a quicker way! We can first simplify the expression inside the logarithm, by factoring out e^x in the numerator:
  (1+e^x)/(1+e^(-x)) = (e^x)(e^(-x)+1)/(1+e^(-x)) = (e^x)(1+e^(-x))/(1+e^(-x)) = e^x.
  Therefore, we have
  d/dx of (ln((1+e^x)/(1+e^(-x)))) = d/dx of (ln(e^x)) = d/dx of (x) = 1. Amazing!
  Comment on Ian Pulizzotto's post “There's even a quicker wa...”
  (51 votes)
George Emmanuel
Posted 4 years ago. Direct link to George Emmanuel's post “should it not be written ...”
should it not be written like this [d/dx(g(x)) .f(x)-
d/d(x)(f(x)).g(x)]/[g(x)]^2, i found this lesson extremely baffling as this opposes what i learned from my textbook, which formula is the right one? any suggestion to improve my understanding,please?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Alex
  Posted 4 years ago. Direct link to Alex's post “Your textbook is incorrec...”
  Your textbook is incorrect; from all other sources, it is indeed f'g - g'f. Try deriving it yourself using the product rule and chain rule.
  Comment on Alex's post “Your textbook is incorrec...”
  (9 votes)
Angelo Luidens
Posted 6 years ago. Direct link to Angelo Luidens's post “In the first Checking You...”
In the first Checking Your Understanding above, isn't the d/dx of e^x = xe^(x-1) ? Else what is the rule applied to e^x?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Howard Bradley
  Posted 6 years ago. Direct link to Howard Bradley's post “The power rule only appli...”
  The power rule only applies when the exponent is a constant. For example x³ or x ⁻².
  It does not apply when the exponent is a variable. For example 2ˣ or, as in this case eˣ.
  
  There's quite a long section on the derivatives of exponential functions, here's the video on eˣ
  https://www.khanacademy.org/math/ap-calculus-ab/ab-derivatives-advanced/ab-diff-exp/v/derivative-of-ex
  Button navigates to signup page
  (3 votes)
Scott Thomas Carter
Posted 5 years ago. Direct link to Scott Thomas Carter's post “What would be the reason ...”
What would be the reason that one couldn't simply take a quotient (for example (x^2 - 3)/x^4 and simply make it the product (x^2-3)*x^-4 and then apply the product rule? Of course, they have different answers because the product rule does not factor in the square of the denominator, but what rule would be broken by doing so?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Ian Pulizzotto
  Posted 5 years ago. Direct link to Ian Pulizzotto's post “Yes, you can express (x^2...”
  Yes, you can express (x^2 - 3)/x^4 as the product (x^2 - 3) * x^-4 and use the product rule to take the derivative. No rule is broken here. Your answer might not appear the same as if you used the quotient rule to differentiate (x^2 - 3)/x^4, but it should end up mathematically equivalent. If not, then you've made an algebraic or differentiation error somewhere.
  
  In fact, the product and chain rules can be used to derive the quotient rule:
  
  d/dx of [f(x)/g(x)] = d/dx of {f(x)*[g(x)]^(-1)}
  = f(x)*(-1)[g(x)]^(-2)*g'(x) + f'(x)*[g(x)]^(-1)
  = [g(x)]^(-2) * [-f(x)*g'(x) + f'(x)*g(x)]
  = [g(x)*f'(x) - f(x)*g'(x)]/[g(x)]^2.
  
  Have a blessed, wonderful day!
  Button navigates to signup page
  (3 votes)
Daleks Incorporated
Posted 5 months ago. Direct link to Daleks Incorporated's post “For the AP test, do you n...”
For the AP test, do you need to fully simplify your answers like you require in the review questions listed here?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
mwood21
Posted 2 years ago. Direct link to mwood21's post “For question 1 I used x^2...”
For question 1 I used x^2 as my f(x) instead of e^x and ended up getting [e^x(2-x)]/(x^3). Would this be incorrect?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Hecretary Bird
  Posted 2 years ago. Direct link to Hecretary Bird's post “Division isn't commutativ...”
  Division isn't commutative like multiplication, so if you switch the positions of the numbers you're dividing, you'll get a different answer. From this, it follows that the derivative of one function divided by a second one would be different than the derivative of the second divided by the first. You don't have to be careful about this when doing the product rule, but when doing the quotient rule, remember that you subtract term with the derivative of the bottom function, and divide by the bottom function squared. Anything else shouldn't give you the right answer, and (e^x (2-x)) / x^3 would be incorrect.
  Button navigates to signup page
  (2 votes)
kevinclauson99
Posted 7 years ago. Direct link to kevinclauson99's post “In problem 2 of example 1...”
In problem 2 of example 1 you state that the answer is

(−xsin(x)+3cos(x))/x^4 however the actual answer is (which the site recognizes) is flip out that middle "+" for a "-"

Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Jesse Cooper
  Posted 7 years ago. Direct link to Jesse Cooper's post “The first minus sign in t...”
  The first minus sign in the solution has been brought out the front of the fraction, so they are stating the answer is −(xsin(x)+3cos(x))/x^4, which is equal to (−xsin(x)-3cos(x))/x^4, rather than (−xsin(x)+3cos(x))/x^4
  Comment on Jesse Cooper's post “The first minus sign in t...”
  (2 votes)
kris kirk
Posted 4 years ago. Direct link to kris kirk's post “so if i had for example '...”
so if i had for example '''y=200x/3+5x^2''' and had to differentiate with respect to y, how would i go about this? ( im pretty certain it's quotient rule) i THINK i have it but i also think i've done it with respect to x though...
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(1 vote)
Answer
- ggadget6
  Posted 4 years ago. Direct link to ggadget6's post “No quotient rule required...”
  No quotient rule required :). You just need the normal derivative rules. Since there are no x's in the denominator, only constants, you can treat 200/3 as a constant, and just use the normal power rule. In this case, your answer would be dy/dx = 200/3 + 10x.
  Button navigates to signup page
  (1 vote)
muhammad muttayab
Posted a year ago. Direct link to muhammad muttayab's post “can rational function be ...”
can rational function be differentiated by doing partial function ? if yes,then which one is more preferable partial fraction or quotient rule?
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(1 vote)
Answer
- Malone Jacoway
  Posted a year ago. Direct link to Malone Jacoway's post “A rational function can b...”
  A rational function can be split into partial fractions before taking the derivative, but this is often a more lengthy process than just doing the quotient rule.
  Button navigates to signup page
  (1 vote)
kaitlynmarushia95
Posted 4 years ago. Direct link to kaitlynmarushia95's post “5x^2/x^2 -9 Why is the 5...”
5x^2/x^2 -9

Why is the 5 put to the side when getting f'(x)
the answer I continue to get is 10x(10x^2 +10x-90)/ (x^2 -9)^2

But the correct answer is -90x/(x^2 -9)^2

I hope someone can help explain this to me.
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(1 vote)
Answer
- Timur Kozhamuratov
  Posted 4 years ago. Direct link to Timur Kozhamuratov's post “If we set 5x^2 = g(x) (fo...”
  If we set 5x^2 = g(x) (for example)
  and x^2-9 = h(x)(for example)
  g'(x) = 5 (2x^2-1)= 10x
  h'(x) = 2x (because h'(x)= 2x^2-1 and 9 =0 according to "power rule",
  which you can check out on Khan academy).
  then f'(x)=g'(x)/h'(x)
  = (g'(x)*h(x)-g(x)*h'(x))/g(x)^2
  substitute (10x)(x^2-9)-(5x^2)(2x)/(x^2-9)^2
  = 10x^3-90x-10x^3/(x^2-9)^2
  = -90x/(x^2-9)^2
  Hope that helps...;)
  Button navigates to signup page
  (1 vote)