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Course: Differential Calculus > Unit 2
Lesson 11: Quotient ruleQuotient rule review
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:
Basically, you take the derivative of
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
Check your understanding
Want to try more problems like this? Check out this exercise.
Example 2
Suppose we are given this table of values:
The Quotient rule tells us that
Check your understanding
Want to try more problems like this? Check out this exercise.
Want to join the conversation?
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?(17 votes)
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!(58 votes)
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?(0 votes)
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.(12 votes)
- 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?(2 votes)
- 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!(7 votes)
For the AP test, do you need to fully simplify your answers like you require in the review questions listed here?(4 votes)- 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?(2 votes)
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(4 votes)
- 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?(1 vote)
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.(3 votes)
- 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...(1 vote)
- 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.(2 votes)
- 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 "-"
(1 vote)
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(2 votes)
- can rational function be differentiated by doing partial function ? if yes,then which one is more preferable partial fraction or quotient rule?(1 vote)
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.(1 vote)
- 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.(1 vote)
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...;)(1 vote)