## 101Ryan101 4 years ago 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.

1. 101Ryan101

I should go study...

2. pk51

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. 101Ryan101

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

4. polpak

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

5. across

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

6. polpak

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

7. 101Ryan101

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

8. polpak

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

9. 101Ryan101

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.

10. polpak

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}$

11. polpak

Yeah

12. 101Ryan101

right on.. foo! :)

13. polpak

I <3 Leibniz notation. Newton can suck it.

14. 101Ryan101

lol