Question

if anyone wants to never ever forget this rule, then here you go with a prove.
(quotient rule for the derivative)

Answer 1

\(\large\color{black}{ \displaystyle y=\frac{\color{green}{f(x)} }{\color{blue}{g(x)}} }\)

starting from a function y, that is equivalent to a quotient of 2 function.


We will use logarithmic differentiation.

\(\large\color{black}{ \displaystyle {\rm Ln}(~y~)={\rm Ln} \left(~\frac{\color{green}{f(x)} }{\color{blue}{g(x)}}~\right) }\)

now, we use: \(\large\color{black}{ \displaystyle {\rm Ln}(~\color{green}{a}~/~\color{blue}{b}~)={\rm Ln}(\color{green}{a})-{\rm Ln}(\color{blue}{b})}\)

\(\large\color{black}{ \displaystyle {\rm Ln}(~y~)= {\rm Ln}(~\color{green}{f(x)}~)-{\rm Ln}(~\color{blue}{g(x)}~) }\)

lets differentiate on both sides, but we need to know, and hope we do, that
\(\large\color{black}{ \displaystyle \frac{d}{dx} {\rm Ln}(~\color{purple}{h(x)}~)=\frac{h'(x)}{h(x)} }\) (for any differentiatable function h(x) )

So lets differentiate the
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\(\large\color{black}{ \displaystyle {\rm Ln}(~y~)= {\rm Ln}(~\color{green}{f(x)}~)-{\rm Ln}(~\color{blue}{g(x)}~) }\)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


\(\large\color{black}{ \displaystyle \frac{dy}{dx}\times\frac{1}{y}= \frac{\color{green}{f'(x)}}{\color{green}{f(x)}}-\frac{\color{blue}{g'(x)}}{\color{blue}{g(x)}} }\)

common denominator,

\(\large\color{black}{ \displaystyle \frac{dy}{dx}\times\frac{1}{y}= \frac{\color{green}{\color{blue}{g(x)}f'(x)}}{\color{green}{\color{blue}{g(x)}f(x)}}-\frac{\color{blue}{\color{green}{f(x)}g'(x)}}{\color{blue}{g(x)\color{green}{f(x)}}} }\)

\(\large\color{black}{ \displaystyle \frac{dy}{dx}\times\frac{1}{y}= \frac{\color{green}{\color{blue}{g(x)}f'(x)}-\color{green}{\color{blue}{g'(x)}f(x)}}{\color{green}{\color{blue}{g(x)}f(x)}} }\)

solving for \(\large\color{black}{ \displaystyle \frac{dy}{dx}}\)

\(\large\color{black}{ \displaystyle \frac{dy}{dx}= y~\times ~~\frac{\color{green}{\color{blue}{g(x)}f'(x)}-\color{green}{\color{blue}{g'(x)}f(x)}}{\color{green}{\color{blue}{g(x)}f(x)}} }\)

we know all the way from the beginning that \(\large\color{black}{ \displaystyle y=\frac{\color{green}{f(x)} }{\color{blue}{g(x)}} }\) so will substitute for y.

\(\large\color{black}{ \displaystyle \frac{dy}{dx}= \frac{\color{green}{f(x)} }{\color{blue}{g(x)}} ~\times ~~\frac{\color{green}{\color{blue}{g(x)}f'(x)}-\color{green}{\color{blue}{g'(x)}f(x)}}{\color{green}{\color{blue}{g(x)}f(x)}} }\)

the \(\large\color{black}{ \displaystyle \color{green}{f(x)} }\)'s cancel, and \(\large\color{black}{ \displaystyle \color{blue}{g(x)} }\) on the bottom becomes squared.


\(\large\color{black}{ \displaystyle \frac{dy}{dx}= \frac{\color{green}{\color{blue}{g(x)}f'(x)}-\color{green}{\color{blue}{g'(x)}f(x)}}{\color{green}{\color{blue}{\left[{\Large\color{white}{|} }g(x)\Large\color{white}{|}\right]^2}}} }\)

Answer 2

(so can be logarithmic differentiation be used to prove the product rule, which would be a bit easier to put in latex)

Answer 3

Oh that's a really clever trick I don't think I've seen anyone do it this way before! I've always just turned the denominator into a negative exponent and sort of did some chain rule/product rule garbage, but I think I like this way you've shown better. It's always nice to see a new look on an old thing.

Answer 4

A nice approach, but it doesn't hold up if \(f(x)=0\) :(

Answer 5

Even so, if \(f(x)=0\) then we know the derivative will be \(0\) because there are other means of figuring that out.

Answer 6

yes.

Answer 7

If the line is y=0, then it;s slope along all the line y=0, is 0,
So if f(x)=0,

then f'(x)=0, f''(x)=0, f'''(x)=0 and on...

Answer 8

and \(\large\color{slate}{ f{'}^{^\infty}(x)=0}\) (excuse me for this notation)

Answer 9

actually, the correct notation is, \(\Large\color{slate}{\displaystyle\frac{d^\infty}{dx^\infty}\left(0\right)=0}\)

Answer 10

I think the notation you're looking for is this: \(\large\color{slate}{ f^{(\infty)}(x)=0}\)

Answer 11

not really, I am saying that any nth derivative of y=0 is going to be zero, not matter how many times you take the derivative of the function....

Answer 12

you can say f^n (0) = 0 ?

Answer 13

where f(x) = 0

Answer 14

I got the notation:) And I was just trying to say that: \(\large\color{slate}{\displaystyle\frac{d^{\infty}}{dx^{\infty}}(0)=0}\)
Saying that anyTH derivative of y=0 is 0.

We can also go through first principles of derivatives for this verification, but i don't see a need.

Answer 15

what happens if f(x) is a negative constant, or g(x) is a negative constant, then ln( f(x)) is undefined

Answer 16

let me think about this one

Answer 17

maybe you can extend the result to complex numbers , that way there isnt a problem

Answer 18

my ln isn't going to be a negativ constant

Answer 19

say you have y = -1 / f(x) , and you want to use the quotient rule on that (obviously it easier to do -f(x)^-1 , but just for arguments sake)

Answer 20

I will take any negative constant out before d/dx

Answer 21

err, y = -1 / g(x) , where g(x) is not zero

Answer 22

I will take constants out

Answer 23

at least negative constants

Answer 24

\(\large\color{black}{ \displaystyle \frac{d}{dx} \left[ -f(x)\right]=-\frac{d}{dx} \left[ f(x)\right] }\)

Answer 25

i am looking at your second step
ln y = ln ( -1 / g(x)) = ln (-1 ) - ln(g(x)

Answer 26

yes, but I will take negative constants out before using logarithmic differentiation.

Answer 27

suppose f(x) = -1, g(x) = x , to be more explicit
y = f(x) / g(x)

Answer 28

that is justa power rule, lol

Answer 29

i know what you are trying to say, although can't think of a case where my prove fails... it does seem like such case most likely exists.

Answer 30

well, for the least part, without negative constants, it is nto bad.

Answer 31

yes its pretty cool. its not a fatal error to have negative constants, you can use
ln(-1) = pi * i

Answer 32

and the derivative of that is just zero

Answer 33

i was thinking of a game

Answer 34

oh, never mind

Answer 35

I thought y=-x^2, reflect across x axis, differentiate (get 2x), and reflect the slope across the x axis, by saying -2x.

which you get either way d/dx (-x^2)=-2x.
but we only need to not have negative constants, not negative coefficients, so it is not necesary

Answer 36

right

Answer 37

actually thats a good question, is ln (-x^2) defined for any x ?

Answer 38

not for any but for x<0

Answer 39

nope, not for x <0 either

Answer 40

well, you can have a case of ln(-x^2) so the quotient rule works with it

Answer 41

since x^2 >= 0

Answer 42

oh, right, not for x<0

Answer 43

lol

Answer 44

but actually

Answer 45

hmm

Answer 46

\(\large\color{slate}{ 2\ln(-x) }\)

Answer 47

no

Answer 48

it is wrong

Answer 49

right

Answer 50

\(\large\color{slate}{ \ln(-x^2) \ne2\ln(-x) }\)

Answer 51

ln(-x^2) = ln(-1) + 2ln(x)

Answer 52

yes, and that we know...

Answer 53

my computer algebra system says
diff( ln(-x^2) , x ) = 2/x ,
so even though the domain of ln(-x^2) is empty on reals it can still differentiate. i think they extend R to C so its valid

Answer 54

mathematica also uses C

Answer 55

This reminds me of a trick for taking the derivative of x^x \[\Large \frac{d}{dx} x^x = x*x^{x-1} + x^x \ln x\] Somehow it magically is the sum of taking the derivative with the power rule and the derivative where you look at x being a base, so to be clear I'm adding the derivative as if \[\Large \frac{d}{dx} x^a = ax^{a-1}\] and \[\Large \frac{d}{dx}b^x = b^x \ln b\] and then substituting in a=x and b=x after I add it.

Answer 56

http://www.wolframalpha.com/input/?i=d%2Fdx+%28ln%28-x%5E2%29%29

Answer 57

this is how wolfram plots y=ln(-x^2)
http://www.wolframalpha.com/input/?i=plot+ln%28-x^2%29

Answer 58

i don't get how it can have a domain, but that is me...

Answer 59

most of the derivative rules in R extend to complex numbers nicely

Answer 60

but, what a calculator closer to my level of math gives
https://www.desmos.com/calculator/x2ij4qov6g

Answer 61

yeah thats what i get with my TI-84

Answer 62

anyway, there aren't many "bad" functions, fortunately.

Answer 63

right :)

Answer 64

I was thinking to make something similar to integration by parts prove from product rule,

but using a quotient rule: \(\large\color{black}{ \displaystyle \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)'f(x)-f(x)g'(x)}{g^2(x)} }\)

\(\large\color{black}{ \displaystyle \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{f'(x)}{g(x)} -\frac{g'(x)f(x)}{g^2(x)}}\)

\(\large\color{black}{ \displaystyle \frac{f(x)}{g(x)}=\int\limits_{~}^{~}\frac{f'(x)}{g(x)}dx~ -~\int\limits_{~}^{~}\frac{g'(x)f(x)}{g^2(x)}dx}\)

\(\large\color{black}{ \displaystyle \frac{f(x)}{g(x)}+\int\limits_{~}^{~}\frac{g'(x)f(x)}{g^2(x)}dx=\int\limits_{~}^{~}\frac{f'(x)}{g(x)}dx}\)

Answer 65

not that i found a case where it can be useful....

Answer 66

no i mean probably not, well it is technically this same integration by parts with a negative powers. As you differentiate the u you have, you will use the chain rule for the u rewriting the u in the denominator as u^-1 .......... i think

Answer 67

yes

Answer 68

yeah , nice problem. just thought of it :)
so the question was, how to take derivative of that, or is it even meaningful since the domain is null

Answer 69

it is treating the domain as R \ {0} since the range will be complex numbers,. Then they graphed the solutions , the real part and imaginary part separately

Answer 70

\(\large\color{slate}{y=\ln(-x^2) }\)

\(\large\color{slate}{y=\log_e(-x^2) }\)

\(\large\color{slate}{e^y=-x^2 }\)

\(\large\color{slate}{y'e^y=-2x }\)

\(\large\color{slate}{y'=-2x/e^x }\)

\(\large\color{slate}{y'=-2x/-x^2 }\)

\(\large\color{slate}{y'=2/x }\)

Answer 71

notice that the imaginary part is a constant Pi

Answer 72

I would rewrite it like this: \[f(x) = \ln(-x^2 ) = \ln ((ix)^2)=2\ln(ix)\]

Answer 73

but that still yields no dom. but you can differentiate though using other means than e^y

Answer 74

I mean, the meaning it has is that you have to just stop thinking with a number line and just get comfortable with a number plane. Seems fine to me lol.

Answer 75

or maybe this
ln(-x^2) = ln(-1) + ln(x^2) = pi*i + ln(x^2)

Real part = ln(x^2)
imaginary part = Pi * i

Answer 76

sure.

Answer 77

i always try to be "real", but I guess I can't always be real. jk

Answer 78

"real" as real numbers

Answer 79

and as real.

Answer 80

now you have the same graph as wolfram
https://www.desmos.com/calculator/jou9jfps5p

Answer 81

:D tnx

Answer 82

I guess I'm not very helpful but I'll say it anyways lol. It just seems like a fixation on "real" numbers is a waste of time. It's like saying "I'll only read books that are printed in black and white with no pictures" when there are a lot of other books we can consider that might help us and make us understand other books written in only black and white with no pictures easier and better.

Answer 83

I like your comparison.... I guess that, though, I would wear a black suit and a white shirt for special occasions.

Answer 84

lol

Answer 85

Oh, latex is back, nice

Answer 86

Haha, well I wear tye dye every day. =P

Answer 87

not that I would always adhere to imaginary numbers, but I agree they're at times very helpful. I guess how helpful they are just depends on a topic.... here, they really were. [tnx for showing me]

Was also going to briefly post, since I used it:
\(\large\color{slate}{ y=\ln(x) }\) re-writing,
\(\large\color{slate}{ e^y=x }\)

differentiating,
\(\large\color{slate}{ y'e^y=1 }\)
\(\large\color{slate}{ y'=1/e^y }\)

we know \(\large\color{slate}{ e^y }\) is \(\large\color{slate}{ x }\)

\(\large\color{slate}{ y'=1/x }\).

this is derivative of ln(x).
(wouldn't do a prove with limit changes)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\(\large\color{slate}{ y=e^x }\)
derivative of \(\large\color{slate}{ e^x }\)

\(\large\color{slate}{ \displaystyle \lim_{h \rightarrow ~0}\frac{f(x+h)-f(x)}{h} }\)

\(\large\color{slate}{ \displaystyle \lim_{h \rightarrow ~0}\frac{e^{x+h}-e^x}{h} }\)

\(\large\color{slate}{ \displaystyle \lim_{h \rightarrow ~0}\frac{e^{x}e^h-e^x}{h} }\)

\(\large\color{slate}{ \displaystyle \lim_{h \rightarrow ~0}\frac{e^{x}(e^h-1)}{h} }\)

\(\large\color{slate}{ \displaystyle \lim_{h \rightarrow ~0}~~~e^{x}\times \frac{(e^h-1)}{h} }\)

\(\large\color{slate}{ \displaystyle \lim_{h \rightarrow ~0}\frac{(e^h-1)}{h} \times e^x }\)

\(\large\color{slate}{ \displaystyle \lim_{h \rightarrow ~0}\frac{(e^h-1)}{h} \times \lim_{h \rightarrow ~0}e^x }\)

https://www.desmos.com/calculator/am9jbwbljd

\(\large\color{slate}{ \displaystyle 1 \times e^x }\)

\(\large\color{slate}{ \displaystyle e^x . }\)

Answer 88

Complex numbers allow you to see some awesome things that you wouldn't see otherwise. For example, \[\Large f(x,y) = x^2-y^2\] Sure, it's a saddle surface but did you know if you travel at a constant radius away from the origin you will exactly trace out a sine curve? \[\Large f(z)=z^2=(x+iy)^2 = x^2-y^2 + i2 xy \\ \Large f(z) = z^2 = (r e^{i \theta})^2 = r^2e^{i2 \theta}=r^2 (\cos 2 \theta + i \sin 2 \theta)\] \[\Large x^2-y^2 = r^2 \cos(2 \theta)\]See, compare the real parts of both and you will see that this naturally extends to any exponent to create the monkey saddle (exponent is 3, has a space for 2 legs and a tail lol) and we could easily get the rotated by 45 degrees version of this same graph \[\Large 2xy = \sin (2 \theta)\] How nice and pretty.

This means if we go on a graph of x^3 from +1 to -1 on the domain but slide on the complex plane we can explain the reason it's negative for odd powers is because it's a sine curve! It sort of brings trigonometry into the idea of (-1)^2n=1 and (-1)^2n+1 =-1 which is fancy.

Answer 89

Note about the complex logarithm, remember: \[\Large z=re^{i (\theta + 2\pi n)}\] So the logarithm is multivalued: \[\Large \ln(z) = \ln(r) + i(\theta + 2\pi n)\] So you still have the original logarithm in there, but the real number line has now become more of a radius and we were only looking at 180 degree rotations before.

If you want to learn more or ask questions I can help you understand this @SolomonZelman

Answer 90

I am not challenging your teaching abilities, rather, I am just saying would that not be too difficult to do?
I do appriciate that you are very willing

Answer 91

Oh yeah it's a lot of fun, complex numbers are easy after you learn several things about them. Have you touched vectors before?

Answer 92

What do you know about complex numbers so far? Maybe you should start a new question since this one is kinda large and OS tends to want to kick me out of stuff that's got a billion questions in it lol.

Answer 93

People seem to be under the impression that they are "imaginary" numbers. Well, all numbers are imaginary last time I checked. I have not been able to look at 5 books on a book shelf and separate the "book" from the "five". I've never even seen a five laying around or anything.

Answer 94

If you consider positive and negative as walking backwards and forwards all complex numbers do is let you walk north, south, east, and west and it has an incredibly convenient notation for doing this. It has a lot of ridiculously beautiful and astounding things to it that made my head spin when I first learned them. The main problem with complex numbers to me was not that I didn't understand what it was saying, it's that they seemed too good to be true.