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### Course: AP®︎/College Calculus AB>Unit 2

Lesson 10: The quotient rule

# Worked example: Quotient rule with table

Let's explore how to find the derivative of F(x) = f(x)/g(x) at x = -1, given the values of f and f' at x = -1 and g(x) = 2x³. By applying the quotient rule, we efficiently calculate the derivative, F'(x), and evaluate it at the specified point.

## 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.
• 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?
• 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.
• 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!!
• 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
• 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.
• Why the Quotient Rule is derived from the Product Rule
• (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
• 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))`?
• 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.
• 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
• 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