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Quotient rule

# Worked example: Quotient rule with table

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.B (LO)
,
FUN‑3.B.2 (EK)
Given the values of f and f' at x=-1 and that g(x)=2x³, Sal evaluates the derivative of F(x)=f(x)/g(x) at x=-1.

## Want to join the conversation?

• How can this rule be derived from the product rule?
• The quotient rule can be derived from the product rule by writing f(x)/g(x) as f(x) * 1/g(x), and using the product, power, and chain rules when differentiating. (Note that 1/g(x) = [g(x)]^(-1).)
d/dx of [f(x)/g(x)] = (d/dx) of [f(x) * 1/g(x)] = f(x) * d/dx of [1/g(x)] + [1/g(x)] * f'(x)
= f(x) * (-1)[g(x)]^(-2) * g'(x) + f'(x)/g(x)
= -f(x)g'(x)/[g(x)]^2 + g(x)f'(x)/[g(x)]^2
= [g(x)f'(x) - f(x)g'(x)] / [g(x)]^2.
• At first, I understood this as f(x)/g(x) * g(x)/g(x). but Why MINUS??
• Basically, the intuition for the quotient rule comes from knowing both the product and chain rules. Here's a video explaining it, in the chain rule section.

However, I am still wondering why this is before the chain rule section. In every video from this point onwards (as far as I know) any time Sal mentions the quotient rule he just says one COULD use it here, but he prefers to just write it was f(x) * g(x)^-1 instead of f(x)/(gx), thus he ends up using the product rule.

Why is this ahead of the chain rule when it really doesn't need to be?

Don't get me wrong, I absolutely love Khan Academy, it is my favourite website ever, I'm just trying to point out some stuff to help them improve.
• I have tried doing this using the product rule instead of the quotient but it just keeps giving me the answer as -2 ! Where am I going wrong?

f'(x) * g^-1 (x) + f(x) * (g^-1 (x))' =
5 * (-2)^-1 + 3 * (6)^-1 =
5/-2 + 3/6 =
-15/6 + 3/6 = -12/6 = -2
• You go wrong when you evaluate (g⁻¹ (x))'.
You need to apply the chain rule because (g⁻¹ (x))' is not simply (g'(x))⁻¹
It is -(g(x))⁻² · g'(x)

So we'd have f'(x) · g⁻¹(x) + f(x) · (-(g(x))⁻² · g'(x)) =
5 · (-2)⁻¹ + 3 · (-(-2)⁻²) · 6)) =
5/-2 - 3·6/4
= -10/4 - 18/4 = -28/4
= -7
• At , could (g(x))^2 be rewritten as g^2(x)? It would look less clunky this way with multiple parentheses, like cos^2(x).
• You could...
but it might get confused with the function composition (g○g)(x), also written as g(g(x)) and - annoyingly for you - g²(x).
I agree it would look less clunky, and that we do it for the trigonometric functions. But, as far as I recall, only for the trig functions. And the trig functions are inconsistent in that sin⁻¹(x) means arcsin(x) rather than 1/sin(x).
So you might further confuse, an already inconsistent "convention". It might be better to write cos²(x) as (cos(x))², but I fear that ship has already sailed....
• Practice ?: e^x/sqrt (x). I get (e^x(2x-1))/(2x^(2/3)). Is the answer: e^x ((sqrt (x)-2 (1/sqrt (x))/x) the same thing? (Sorry about the confusing formatting). Thank you!!
(1 vote)
• The answer is not the same as what you got. I will work it out and you can check your work:
First, I notice that this a differentiation requiring the quotient rule:
``[f'(x)g(x)-f(x)g'(x)]/(g(x))^2``

Where
`` f(x) = e^x, and f'(x) = e^x``

and
`` g(x) = sqrt(x) or x^(1/2) and g'(x) = 1/(2sqrt(x)) or 1/2(x^(-1/2))``

Therefore:
The derivative is
`` [e^x * sqrt(x) - e^x(1/2)x^(-1/2)]/x ``

Simplified it is
`` [e^x*(2x-1)]/(2x^(3/2)``

It looks like you only made a mistake when simplifying the power of x in the denominator
• Curious...is there any marked advantage to using the d/dx[f(x)] versus just f'(x)? Is d/dx the proper notation and the prime notation just a less formal definition of it?
If they're interchangeable, why use a significantly more complex notation instead of the simpler, faster notation?
(1 vote)
• d/dx indicates that you're differentiating with respect to x, which may be important to specify if your function has multiple variables and constants in it. But when there is no ambiguity, the f'(x) notation is more compact and often easier to typeset.
• Why the Quotient Rule is derived from the Product Rule
(1 vote)
• (f(x)/g(x))'
= (f(x)*g(x)^(-1))'
= f'(x)*g(x)^(-1) + f(x)*(g(x)^(-1))'
= f'(x)*g(x)^(-1) - f(x)*g'(x)*g(x)^(-2)
= (f'(x)*g(x) - f(x)*g'(x)) * g(x)^(-2)
= (f'(x)*g(x) - f(x)*g'(x)) / g(x)^2
• Shouldn't the title of the video be
Worked example: Quotient Rule with mixed implicit and explicit?
It's the same format as the one in the Product Rule section, and there are near to no mentions of tables in this video.

The thing is, though, the practice problems all actually have tables, unlike the video.
(1 vote)
• Wouldn't it be easier just to evaluate `d/dx[f(x)/g(x)]` as `f(x)*(1/g'(x))+f'(x)*(1/g(x))`?
(1 vote)
• The first term would be f(x)•(1/g(x))', not f(x)•(1/g'(x)). And computing (1/g(x))' still required quotient rule or chain rule.
(1 vote)
• hello, I have tried to solve this question in the quotient rule way but the answer is different from what I solve. this is the problem
d\dx (5x^2-3x)/(4x)

and this is the solution
5/4

I don't get it
(1 vote)
• d/dx(5x^2-3x)/4x
[d/dx(5x^2-3x)•4x-d/dx(4x)•(5x^2-3x)]/(4x)^2 Quotient Rule
[(10x-3)•4x-4•(5x^2-3x)]/(16x^2) compute derivatives
[40x^2-12x-20x^2+12x]/(16x^2) distribute
[20x^2]/(16x^2) combine like terms
5/4 cancel terms

Alternatively, we can compute this without quotient rule:

d/dx(5x^2-3x)/(4x)
d/dx(5x/4-3/4)
5/4
(1 vote)

## Video transcript

- [Voiceover] Let f be a function such that f of negative one is equal to three, f prime of negative one is equal to five. Let g be the function g of x is equal to two x to the third power. Let capital F be a function defined as, so capital F is defined as lowercase f of x divided by lowercase g of x, and they want us to evaluate the derivative of capital F at x equals negative one. So the way that we can do that is, let's just take the derivative of capital F, and then evaluate it at x equals one. And the way they've set up capital F, this function definition, we can see that it is a quotient of two functions. So if we want to take it's derivative, you might say, well, maybe the quotient rule is important here. And I'll always give you my aside. The quotient rule, I'm gonna state it right now, it could be useful to know it, but in case you ever forget it, you can derive it pretty quickly from the product rule, and if you know it, the chain rule combined, you can get the quotient rule pretty quick. But let me just state the quotient rule right now. So if you have some function defined as some function in the numerator divided by some function in the denominator, we can say its derivative, and this is really just a restatement of the quotient rule, its derivative is going to be the derivative of the function of the numerator, so d, dx, f of x, times the function in the denominator, so times g of x, minus the function in the numerator, minus f of x, not taking its derivative, times the derivative in the function of the denominator, d, dx, g of x, all of that over, so all of this is going to be over the function in the denominator squared. So this g of x squared, g of x, g of x squared. And you can use different types of notation here. You could say, instead of writing this with a derivative operator, you could say this is the same thing as g prime of x, and likewise, you could say, well that is the same thing as f prime of x. And so now we just want to evaluate this thing, and you might say, wait, how do I evaluate this thing? Well, let's just try it. Let's just say we want to evaluate F prime when x is equal to negative one. So we can write F prime of negative one is equal to, well everywhere we see an x, let's put a negative one here. It's going to be f prime of negative one, lowercase f prime, that's a little confusing, lowercase f prime of negative one times g of negative one, g of negative one minus f of negative one times g prime of negative one. All of that over, we'll do that in the same color, so take my color seriously. Alright, all of that over g of negative one squared. Now can we figure out what F prime of negative one f of negative one, g of negative one, and g prime of negative one, what they are? Well some of them, they tell us outright. They tell us f and f prime at negative one, and for g, we can actually solve for those. So, let's see, if this is, let's first evaluate g of negative one. G of negative one is going to be two times negative one to the third power. Well negative one to the third power is just negative one, times two, so this is negative two, and g prime of x, I'll do it here, g prime of x. Let's use the power rule, bring that three out front, three times two is six, x, decrement that exponent, three minus one is two, and so g prime of negative one is equal to six times negative one squared. Well negative one squared is just one, so this is going to be equal to six. So we actually know what all of these values are now. We know, so first we wanna figure out f prime of negative one. Well they tell us that right over here. F prime of negative one is equal to five. So that is five. G of negative one, well we figured that right here. G of negative one is negative two. So this is negative two. F of negative one, so f of negative one, they tell us that right over there. That is equal to three. And then g prime of negative one, just circle it in this green color, g prime of negative one, we figured it out. It is equal to six. So this is equal to six. And then finally, g of negative one right over here. We already figured that out. That was equal to negative two. So this is all going to simplify to... So you have five times negative two, which is negative 10, minus three times six, which is 18, all of that over negative two squared. Well negative two squared is just going to be positive four. So this is going to be equal to negative 28 over positive four, which is equal to negative seven. And there you have it. It looks intimidating at first, but just say, okay, look. I can use the quotient rule right over here, and then once I apply the quotient rule, I can actually just directly figure out what g of negative one, g prime of negative one, and they gave us f of negative one, f prime of negative one, so hopefully you find that helpful.