Question

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.

Answer 1

I should go study...

Answer 2

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.

Answer 3

when you're doing related rate problems though that extra dx/dt... bllllaaaaa.. :(

Answer 4

y is the dependent variable. y' = dy/dx = d/dx of y

Answer 5

The notation dy/dx is very useful when integrating and it can simply be read as "the derivative of y with respect to x."

Answer 6

you don't get the dx/dy when you use leibniz notation taking the differential of that equation.

Answer 7

I get confused when we start doing related rates and dt is thrown in the mix. that what's going on.

Answer 8

\[y = x^3\]\[\implies \frac{d}{dx}[y] = \frac{d}{dx}[x^3]\]\[\implies \frac{dy}{dx} = 3x^2\]

Answer 9

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.

Answer 10

If however we wanted to take the differential with respect to t:

\[\frac{d}{dt}[y] = \frac{d}{dt}[x^3]\]\[\implies \frac{dy}{dt} = \frac{d}{dx}[x^3] \cdot \frac{dx}{dt}\]\[\implies \frac{dy}{dt} = 3x^2\frac{dx}{dt}\]

Answer 11

Yeah

Answer 12

right on.. foo! :)

Answer 13

I <3 Leibniz notation. Newton can suck it.

Answer 14

lol