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101Ryan101
Group Title
f(x) = x^3
f'(x) = 3x^2
why then when we use this notation:
y = x^3
dy/dx = 3x^2(dx/dy) <isn't y the dependent variable.. I'm so confused today.
 3 years ago
 3 years ago
101Ryan101 Group Title
f(x) = x^3 f'(x) = 3x^2 why then when we use this notation: y = x^3 dy/dx = 3x^2(dx/dy) <isn't y the dependent variable.. I'm so confused today.
 3 years ago
 3 years ago

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101Ryan101 Group TitleBest ResponseYou've already chosen the best response.0
I should go study...
 3 years ago

pk51 Group TitleBest ResponseYou've already chosen the best response.1
Two peple developed Calculus, Newton and Leibniz. Newton chose to use primes "'" , or f'(x) while lebinz chose dy/dx, the differential in y/the differential in x. They both mean the same thing for now, but dy/dx becomes important later on in Calculus 2 and 3.
 3 years ago

101Ryan101 Group TitleBest ResponseYou've already chosen the best response.0
when you're doing related rate problems though that extra dx/dt... bllllaaaaa.. :(
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.2
y is the dependent variable. y' = dy/dx = d/dx of y
 3 years ago

across Group TitleBest ResponseYou've already chosen the best response.0
The notation dy/dx is very useful when integrating and it can simply be read as "the derivative of y with respect to x."
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.2
you don't get the dx/dy when you use leibniz notation taking the differential of that equation.
 3 years ago

101Ryan101 Group TitleBest ResponseYou've already chosen the best response.0
I get confused when we start doing related rates and dt is thrown in the mix. that what's going on.
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.2
\[y = x^3\]\[\implies \frac{d}{dx}[y] = \frac{d}{dx}[x^3]\]\[\implies \frac{dy}{dx} = 3x^2\]
 3 years ago

101Ryan101 Group TitleBest ResponseYou've already chosen the best response.0
Ok, and so if I took both sides with respect to dt... that's why there's an "extra" dx/dt.. Ok... I get it.
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.2
If however we wanted to take the differential with respect to t: \[\frac{d}{dt}[y] = \frac{d}{dt}[x^3]\]\[\implies \frac{dy}{dt} = \frac{d}{dx}[x^3] \cdot \frac{dx}{dt}\]\[\implies \frac{dy}{dt} = 3x^2\frac{dx}{dt}\]
 3 years ago

101Ryan101 Group TitleBest ResponseYou've already chosen the best response.0
right on.. foo! :)
 3 years ago

polpak Group TitleBest ResponseYou've already chosen the best response.2
I <3 Leibniz notation. Newton can suck it.
 3 years ago
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