/sci/ - Science & Math

[Anonymous No.8960492]

>taking calc II
>prof spends first day reviewing calc I
>points out that instead of using quotient rule you can more quickly find f' by taking ln of both sides and differentiating implicitly
>tfw knowing properties of logs and differentiation rules should have been able to figure this out myself
>feel like a brainlet

[Anonymous No.8960499]

>>8960492
>taking ln of both sides and differentiating implicitly

That is a horrible suggestion.

[Anonymous No.8960520]

>>8960492
This. >>8960499

How is the quotient rule even difficult? If rational functions scare you, just write the denominator raised to -1 and use the product rule (my specialty).

[Anonymous No.8960522]

Is this bait?
6/10

[Anonymous No.8960527]

Im sorry for you, but was this apart of a full lecture or just a method he suggested?

[Anonymous No.8960535]

>>8960527
Just a method he suggested for when you have products nested in the top/bottom.

[Anonymous No.8960570]

>>8960492
Calc 3 fag here. Nobody ever told me this nor have I figured it out on my own.

That said, is there any case in which this is actually useful?

[math] f(x) = \frac{x}{ln(x)} \\ ln(f(x)) = ln(x) - ln(ln(x)) \\ \frac{f'(x)}{f(x)} = \frac{1}{x} - \frac{1}{x ln(x)} \\ f'(x) = \frac{\frac{x}{ln(x)}}{x} - \frac{\frac{x}{ln(x)}}{x ln(x)} [/math]

Looks to me like it would be simply to apply the formula, because if you have used it plenty you can just recite it without thinking. Meanwhile in this method you are forced to do algebra.

[Anonymous No.8960579]

>>8960570
I've seen this before.
I think it's only useful in incredibly rare situations when differentiating exponential functions, but that's it.

[. No.8960617]

>>8960492 >>8960499 >>8960520 >>8960570

There is a nice method found in Richard Feynman's book "Tips on Physics" Page 22

Let [math] f = k \cdot u^a \cdot v^b \cdot w^c \dots [/math] a function with respect to t

[math] \frac{df}{dt} = f \left( \frac{a}{u} \frac{du}{dt} + \frac{b}{v} \frac{dv}{dt} + \frac{c}{w} \frac{dw}{dt} + \dots \right) [/math]

where k, a, b, c, ... are constants

i.e.

Let [math] f = k \prod_{i=1}^{n} u_{i} \^ a_{i} [/math]

[math] \frac{df}{dt} = f \sum_{i=1}^{n} \frac{a_i}{u_i} \frac{du_i}{dt} [/math]

>link of the source here: https://archive.org/stream/FeynmanTipsOnPhysics/Feynman-Tips-on-Physics#page/n35/mode/2up

[. No.8960621]

>>8960492 >>8960499 >>8960520 >>8960570

There is a nice method found in Richard Feynman's book "Tips on Physics" Page 22

Let [math] f = k \cdot u^a \cdot v^b \cdot w^c \dots [/math] a function with respect to t

[math] \frac{df}{dt} = f \left( \frac{a}{u} \frac{du}{dt} + \frac{b}{v} \frac{dv}{dt} + \frac{c}{w} \frac{dw}{dt} + \dots \right) [/math]

where k, a, b, c, ... are constants

i.e.

Let [math] f = k \prod_{i=1}^{n} u_{i} \^{} a_{i} [/math]

[math] \frac{df}{dt} = f \sum_{i=1}^{n} \frac{a_i}{u_i} \frac{du_i}{dt} [/math]

>link of the source here: https://archive.org/stream/FeynmanTipsOnPhysics/Feynman-Tips-on-Physics#page/n35/mode/2up

[. No.8960626]

>>8960617 >>8960621

Oh God! I messed up latex again. I'll try one more time.

>>8960492 >>8960499 >>8960520 >>8960570

There is a nice method found in Richard Feynman's book "Tips on Physics" Page 22

Let [math] f = k \cdot u^a \cdot v^b \cdot w^c \dots [/math] a function with respect to t

[math] \frac{df}{dt} = f \left( \frac{a}{u} \frac{du}{dt} + \frac{b}{v} \frac{dv}{dt} + \frac{c}{w} \frac{dw}{dt} + \dots \right) [/math]

where k, a, b, c, ... are constants

i.e.

Let [math] f = k \prod_{i=1}^{n} u_{i} [/math] ^ [math] a_{i} [/math]

[math] \frac{df}{dt} = f \sum_{i=1}^{n} \frac{a_i}{u_i} \frac{du_i}{dt} [/math]

>link of the source here: https://archive.org/stream/FeynmanTipsOnPhysics/Feynman-Tips-on-Physics#page/n35/mode/2up

[Anonymous No.8960648]

>>8960492

Taking the natural log of an expression to change its operations is quite useful in some cases. Here's the most elementary example I can think of,

[math]\ln \prod_{k=1}^N f_k [/math]

gives you

[math] \sum_{k=1}^N f_k [/math]

It's trivial stuff, but it's a way to ease you into this kinda crap.

[Anonymous No.8960654]

>>8960492
pretty sure that technique is in Stewart's chapter 3 or something.

[Anonymous No.8960828]

>>8960520
>my specialty

[Anonymous No.8960846]

>>8960492
>>points out that instead of using quotient rule you can more quickly find f' by taking ln of both sides and differentiating implicitly

Wut?
y=f(x)/g(x)
ln(y)=ln(f(x)) - ln(g(x))
y'/y = f'(x)/f(x) - g'(x)/g(x)
y' = f'(x)/g(x) - g'(x)f(x)/g^2(x)

Ok, it works (if f(x)=/=0) but just use the product rule ffs:
y=f(x)/g(x)
y'=f'(x)/g(x) - f(x)g'(x)/g^2(x)

One step and you're done.

[Anonymous No.8960880]

>>8960520
>specialty
Kek

[Anonymous No.8960885]

>>8960492
This is the first thing my class went over when we did implicit differentiation.
[eqn]\frac{d}{dx}y = \frac{y'}{y}[/eqn] when [math]y[/math] is dependent on [math]x[/math]. Multiply [math]y[/math] over and you get [math]y'[/math]

[Anonymous No.8960887]

>>8960885
oops meant [eqn]\frac{d}{dx}\ln \left(y\right)[/eqn]

[Anonymous No.8960910]

>>8960492
If you wanted to differentiate [math]y = x^x[/math]
Take natural log of both sides [math]\ln y=x\ln x[/math]
Differentiate both sides
[math]\frac{y'}{y} = 1 + \ln x[/math]
Multiply y over
[math]y' = y(1+\ln x)[/math]
Sub y for x^x
[math]y' = x^x(1+\ln x)[/math]

[Anonymous No.8961040]

>>8960520
I always thought this was just common sense, especially if the denominator is already some function to a power.

[Anonymous No.8961045]

>tfw genius
>instantly compute any polinomial derivative in my brain

weak ass niggas

[Anonymous No.8961273]

>>8961045
Pffft

Couple weeks into calculus 1 now, doing well, already past the chain rule and beyond. Quotient rule was a joke. Product rule remains my specialty.

[Anonymous No.8961329]

>tfw taking calc II a second time
>At the start of the year the teacher said he had to retake it

[Anonymous No.8961662]

>>8960522
I hope so
>>8960492
quotient rule is easy, why would anyone consider not doing it if they can? Sure there might be another way to do it, but you run the risk of looking a right autist.

[Anonymous No.8961687]

>>8960492
prof is trolling. he wants kids to try and fail on the test.

[Anonymous No.8961711]

>>8960846
It works only if f, g >=0

[Anonymous No.8962087]

>>8960828
>>8960880
Newfags?

[Anonymous No.8962196]

>>8962087
It just never gets old

[Anonymous No.8962229]

Isn't this high school math? What the fuck do they teach at school?

[Anonymous No.8962248]

>>8961687
any more of these?

[Anonymous No.8962836]

>>8962196
underage, then..

[Anonymous No.8962843]

>>8961711
g=/=0 come from the problem. Using logs adds an extra requirement that f=/=0. f and g can still be negative.

[Anonymous No.8963024]

>>8960626
Pretty neat. I can't really think of any time I would have needed it though.

[Anonymous No.8963349]

>>8962836
Nah. Couple of weeks into calculus 1 now, doing well

[Anonymous No.8963403]

>>8962836
haha, right, surely nobody over 18 can appreciate humor!

[Anonymous No.8963425]

This seems like a bunch of extra steps.....
Just use the product rule.

y = f/g = f(g^-1)
y' = f'g^-1 + f(g^-1)'

Then again the quotient rule isn't that hard but at the lest with the product rule if you change the order you get the same answer but if you mix up the order of the quotient rule you get the wrong answer.

Anyways it sounds like you're just starting a summer course and you better do your fucking homework. Summer courses are compressed as fuck and if you get behind or don't understand something you're going to fucking fail. Like really fail. Not a B, not a C+, but a big fuckin F.

[Anonymous No.8963449]

>>8963425
This. Calculus 2 is fast as fuck anyway. You'd have to be insane or already know the material and just taking it for the credit to take this class in the summer.

[Anonymous No.8963549]

>>8960520
yeah sometimes i wonder why the quotient rule even exists