What does dy "with respect to" dx mean?

[Lime]

I don't get the whole "with respect to" thing.

As an example, what would dy with respect to dx mean? I know it's written as dy/dx.

 

[galactus]

Suppose you had \(\displaystyle y=x^{2}+2x-1\)

If you differentiated, you'd have \(\displaystyle \frac{dy}{dx}=2x+2\)

\(\displaystyle dy=(2x+2)dx\)

y with respect to x.

 

[daon]

I don't get the whole "with respect to" thing.

As an example, what would dy with respect to dx mean? I know it's written as dy/dx.

If you have some function y=f(x), by writing \(\displaystyle \frac{dy}{dx}=...\)

You should be asking yourself "as x changes, what is happening to y?" and when you solve for \(\displaystyle \frac{dy}{dx}\) you are calculating just that. The instantaneous change in y when x changes.

So, given y=2x, you calculate dy/dx to be 2. Thus when x changes by any amount, y will change twice as much (or fast if y is a function of time)!

Alternatively, it may be necessary/helpful to consider \(\displaystyle \frac{dx}{dy}\). Although, generally, y is a function of x, in our example we can alsy say x is a function of y since y=2x is 1-1. So, taking the derivative of y=2x with respect to y, we get \(\displaystyle \frac{dx}{dy}=\frac{1}{2}\). SO, when y changes by any amount, x will change half as much.