4 ms·

d/dx is not a fraction but an operation, dy/dx is a fraction

• rtpg 3y ago
  What series of operations can lead you to end up with dy/dx though? Like Integral{x^2){dx} is how you're introducing dx into a formula, but it's not like integration just works over division, so I'm a bit stuck on the construction phase
  • lupire 3y ago
    y = x^2 dy = 2x dx dy/dx = 2x Chain rule makes this not completely trivial formulation.
• xvedejas 3y ago
  Is `dy/dx` ever distinct from `d/dx y`? Generally I've treated `df(x)/dx` as just another way to write the operator `d/dx` applied to `f(x)`, which I don't think of as a fraction.
  • lupire 3y ago
    dy/dt = dy/dx * dx/dt
    • xvedejas 3y ago
      That's just the chain rule, right? d/dt y(t) = ( d/dx y(x) )( d/dt x(t) )