[AtomicX (OP)]

Hello, I have some questions considering derivatives because i am new in calculus. So I was wondering if we had an equation like this : dy/dx = dy * k (the derivative of y with respect to x is equal to dy times k) is it legit to divide with dy and end up with k = 1/dx? and also if we have this : dx = dp can we divide both sides with dy and get dx/dy = dp/dy (the derivative of x with respect to y is equal to the derivative of p with respect to y) and vice-versa can we multiply with dy to get dx = dp ? All these might be very obvious but I am asking to clear any misunderstandings I have.

[alancalverd]

Not strictly true. You need to distinguish between Δy, a small increment in y, and dy/dx, the rate of change of y with respect to x.

If y is a distance and x is time, Δy is still a distance (meters) but dy/dx is a speed (meters per second), so the meaning of dy * k is not obvious, whereas Δy * k may well be meaningful.

Consider a staircase. It's made from horizontal (Δx) and vertical (Δy) segments. Suppose for simplicity that they are equal, 10 cm lengths. Δy/Δx = 1, obviously.
I want to rise 10 m.
To make 100 steps you need 2 x 10 x 100 = 2000 cm of wood.
But to make a 45-degree ramp (dy/dx = 1)  of the same height, you only need 1414 cm of wood.

[AtomicX (OP)]

Thanks for the answer but, do you mean that dy can only be used in a derivative like dy/dx ? and furthermore you said the meaning of dy * k = something, is not obvious but what if we divide it by dt e.g. so it becomes speed and it has a meaning, is that ok ?  ([dy/dt] * k = something / dt).

[chris]

Not strictly true. You need to distinguish between Δy, a small increment in y, and dy/dx, the rate of change of y with respect to x.

If y is a distance and x is time, Δy is still a distance (meters) but dy/dx is a speed (meters per second), so the meaning of dy * k is not obvious, whereas Δy * k may well be meaningful.

Consider a staircase. It made from horizontal (Δx) and vertical (Δy) segments. Suppose for simplicity that they are equal, 10 cm lengths. Δy/Δx = 1, obviously.
I want to rise 10 m.
To make 100 steps you need 2 x 10 x 100 = 2000 cm of wood.
But to make a 45-degree ramp (dy/dx = 1)  of the same height, you only need 1414 cm of wood.

You should have been my maths / physics teacher: then I'd have actually understood all this the first time round and not had to re-teach myself later...

[alancalverd]

If dy has a meaning, it is "the limiting infinitesimal". dy/dx is defined as Limit_Δx→0 Δy/Δx, and it is quite clear from the staircase example that a truly smooth ("differentiable") function is different from a stepwise approximation. You could argue that the limiting value of Δy being zero, dy * k is always zero, so not a useful concept.

When your studies progress from differentials to integrals, you will come across the form ∫(function)dx. This doesn't mean the integral of "function multiplied by dx",  but of "function developed smoothly along the x axis". IIRC the Open University and several other modern authorities decry that notation, preferring something along the lines of
∫_x(function).

I was brought up under the "old" notation but would recommend the OU style wherever the meaning is unambiguous. Interestingly, the OU teaches integrals before differentials, on the grounds that the area under a curve (the integral of a 2-dimensional function) is easier to visualise and understand than the slope of a curve (the differential)

[alancalverd]

You should have been my maths / physics teacher: then I'd have actually understood all this the first time round and not had to re-teach myself later...

Thanks for the compliment, but it took me around 20 years after A levels (with lots of distinctions) to actually understand what I had written in the exams! My maths teacher insisted that engineers used formulae, but mathematicians dreamed about them.

[alancalverd]

what if we divide it by dt e.g. so it becomes speed and it has a meaning, is that ok ?  ([dy/dt] * k = something / dt).

If dy/dt = 100 meters/second and k = 2, then [dy/dt]*k = 200 m/s.

You can differentiate again: d²y/dt² is the rate of change of dy/dt, i.e. acceleration, (meters per second) per second. The precise notation is important and consistent with the abbreviation m/s²