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101Ryan101

  • 3 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.

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  1. 101Ryan101
    • 3 years ago
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    I should go study...

  2. pk51
    • 3 years ago
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    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
    • 3 years ago
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    when you're doing related rate problems though that extra dx/dt... bllllaaaaa.. :(

  4. polpak
    • 3 years ago
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    y is the dependent variable. y' = dy/dx = d/dx of y

  5. across
    • 3 years ago
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    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
    • 3 years ago
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    you don't get the dx/dy when you use leibniz notation taking the differential of that equation.

  7. 101Ryan101
    • 3 years ago
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    I get confused when we start doing related rates and dt is thrown in the mix. that what's going on.

  8. polpak
    • 3 years ago
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    \[y = x^3\]\[\implies \frac{d}{dx}[y] = \frac{d}{dx}[x^3]\]\[\implies \frac{dy}{dx} = 3x^2\]

  9. 101Ryan101
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    Yeah

  12. 101Ryan101
    • 3 years ago
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    right on.. foo! :)

  13. polpak
    • 3 years ago
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    I <3 Leibniz notation. Newton can suck it.

  14. 101Ryan101
    • 3 years ago
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    lol

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