What does dy "with respect to" dx mean?

Lime

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Sep 8, 2006
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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.
 
Suppose you had y=x2+2x1\displaystyle y=x^{2}+2x-1

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

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

y with respect to x.
 
Lime said:
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 dydx=...\displaystyle \frac{dy}{dx}=...

You should be asking yourself "as x changes, what is happening to y?" and when you solve for dydx\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 dxdy\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 dxdy=12\displaystyle \frac{dx}{dy}=\frac{1}{2}. SO, when y changes by any amount, x will change half as much.
 
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