I Question about the quotient rule of derivatives

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A quick question about the theory behind the quotient rule?
Now, I understand how to use the quotient rule for derivatives and everything. I do not struggle with using it, my question is mostly about the formula itself...I very much enjoy WHY we do things in math, not just “here’s the formula, do it”....Here is the formula for the quotient rule of derivatives.

A688273F-7DB7-447F-9B43-B63AADA0DA65.jpeg

Now, my question is. Why do we square the g(x) in the denominator? I almost feel like the formula for quotient rule should just be what is in the numerator? Why do we square the g(x)? Where does that come from? Why g(x) squared?
 

phyzguy

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Do you know the product rule and the chain rule? Why don't you try deriving the quotient rule using the product rule and the chain rule?
 
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Do you know the product rule and the chain rule? Why don't you try deriving the quotient rule using the product rule and the chain rule?
would that explain why the g(x) is squared? It just seems weird to me. Why not just have g(x) function in the denom.?The fact that it is squared makes me wonder it’s origin.
 

fresh_42

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would that explain why the g(x) is squared? It just seems weird to me. Why not just have g(x) function in the denom.?The fact that it is squared makes me wonder it’s origin.
It is squared because ##x^n## differentiates to ## \sim x^{n-1}## and with ##n=-1## you get the one square on the right. You haven't answered @phyzguy 's question!
 

Delta2

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Maybe its not a good idea to spoon feed you but sometimes there is no other way, here it is, read this Wikipedia article with 3 different proofs of the quotient rule that will help you understand the "inner mechanisms" and the ultimate why's, pick the one you like.
 

RPinPA

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I always have trouble remembering the order of the terms in the numerator, so I tend to use the product rule.

As people have been suggesting, try using the product rule on ##f(x) [g(x)]^{-1}## and you should see exactly where all of the terms come from. Don't take our word for it, it will really help your understanding.
 

fresh_42

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I always have trouble remembering the order of the terms in the numerator, so I tend to use the product rule.
And I always thought I was the only one ...
 

phyzguy

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And I always thought I was the only one ...
Me too! I realized early on that the quotient rule was just a consequence of the product rule, so I didn't need to memorize the quotient rule. I never use it.
 
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I always have trouble remembering the order of the terms in the numerator, so I tend to use the product rule.
And I always thought I was the only one ...
I came up with my own mnemonic device for the quotient rule, and one I've never seen anywhere else. Here it is, in the context of the differential of u/v.
$$d(\frac u v) = \frac{v du - u dv}{v^2}$$
How do I remember which term in the numerator comes first? The vd one, an abbreviation for something unrelated to mathematics.

For a derivative instead of a differential, replace du by du/dx or u' and similar for dv.
 

HallsofIvy

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The derivative of [itex]\frac{f(x)}{g(x)}[/itex] is given, using the definition of the derivative, by [itex]\lim_{h\to 0}\frac{\frac{f(x+h)}{g(x+h)}- \frac{f(x)}{g(x)}}{h}[/itex].
To do that quotient, [itex]\frac{f(x+h)}{g(x+h)}- \frac{f(x)}{g(x)}[/itex], get the "common denominator", [itex]g(x)g(x+ h)[/itex]: [itex]\frac{f(x+ h)g(x)}{g(x)g(x+h)}- \frac{f(x)g(x+h)}{g(x)g(x+h)}[/itex]. (it is that "g(x)g(x+h)" in the denominator that will give "[itex]g^2(x)[/itex]" after we take the limit.)
 

symbolipoint

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I came up with my own mnemonic device for the quotient rule, and one I've never seen anywhere else. Here it is, in the context of the differential of u/v.
$$d(\frac u v) = \frac{v du - u dv}{v^2}$$
How do I remember which term in the numerator comes first? The vd one, an abbreviation for something unrelated to mathematics.

For a derivative instead of a differential, replace du by du/dx or u' and similar for dv.
Another way is like this:
Function (Hi)/(Ho)

Mnemonic: Ho d Hi minus Hi d Ho, over Ho Ho

Symbolified: (Ho*dHi-Hi*dHo)/(HoHo)
(Try writing on paper using better 'typesetting' to see it better.)
 

Stephen Tashi

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would that explain why the g(x) is squared?
Yes, if you know how to apply the chain rule to differentiate ##(g(x))^{-1}##.

I find the explanation given by @HallsofIvy the most intuitive explanation of the denominator in the quotient rule. To explain the numerator, you need the same trick used to prove the product rule.
 

FactChecker

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The Quotient Rule Rhyme (for D hi/low):
If the quotient rule you wish to know,
It's low Dhi less hi Dlow
Draw a line and down below
The denominator squared must go.
 

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