>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
Taking calc II
Brody Lee
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Jonathan Moore
>taking ln of both sides and differentiating implicitly
That is a horrible suggestion.
Justin Clark
This. 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).
Josiah Howard
Is this bait?
6/10
Ian Rogers
Im sorry for you, but was this apart of a full lecture or just a method he suggested?
Daniel Young
Just a method he suggested for when you have products nested in the top/bottom.
Samuel Howard
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.
Lincoln Peterson
I've seen this before.
I think it's only useful in incredibly rare situations when differentiating exponential functions, but that's it.
Hunter Sanders
Oh God! I messed up latex again. I'll try one more time.
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: archive.org
Henry Bell
pretty sure that technique is in Stewart's chapter 3 or something.