Derivitive of X with Respect to Time

  • Thread starter bmed90
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So I just recently got into calculus but there is one thing that keeps confusing me, and that is the term "with respect to (whatever)". Heres what I register, take the derivative of whatever equation you have. So for example dx/dt, I register take the derivitive of all the terms with X.

Like for example X2+1

would be 2x

But then you says if you say, with respect to time. So like, I know as time is changing to is the function, because maybe all the x values are changing as time continues. Like a car driving on the interstate. But the term "respect to whatever" just confuses me for some reason. I dont know why. It throws me off in lecture when my physics teacher reffers to calculus equations to relate physics equations and all that. Maybe I'm being close minded, I dont know. But if you guys could give me a solid piece of advice on what to make of the whole "with respect to whatver" stuff it would be great. Im ready to rock on and keep it rockin. cool
 

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  • #2
CompuChip
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You can also differentiate x2 + 1 with respect to time. As you said, when x is a function of time itself (x = x(t), like in a physics problem) this gives a different result (you need something called the chain rule, which I won't go into here).

Here's another (more mathematical) example. Suppose that you have a function of several variables, for example, f(x, y) = x2 + y2. Now let's forget about the "with respect to" part and consider the question: "what is the derivative of f"?
Well, this doesn't make a lot of sense. For just one variable, f(x), the derivative f'(x) would tell you something about the slope of f, in other words: about the change in f(x) when you vary x a little bit. But in f(x, y), we have two variables what we can vary. So the question should be reformulated as: "what is the change in f, if we vary x (or y) a little bit?"*). So instead of: "what is the derivative of f?" we must ask, "what is the derivative of f with respect to x (or y)?".
To do the calculation, you only need to know differentiation in one variable: you just consider all the others as constants. So for example,
df/dx = 2x,
df/dy = 2y
because in the first one, the derivative of y2 with respect to x is zero (just like the derivative of, say, 24 would) and similarly for x2 in the second one.
In the same way, if g(x, y) = 2x + xy - y3, you would have
dg/dx = 2 + y
dg/dy = x - 3 y2.

Of course you can always draw f(x, y) (or g(x, y)) in three dimensions (draw the value of f(x,y) on the z-axis above the point (x, y) in the horizontal plane).
To visualise this in terms of slope, you could then say: df/dx is the slope of the graph which I get, when intersecting this graph with the (y, z) plane. In other words, it is the slope of the graphs hy(x) where y is just some constant number, and h is just a function of x.

*) There are other questions, like: "what happens to f if we vary (x, y) a little bit, say in the direction (a, b)" or "what is the largest possible change of f for a variation of (x, y) in any direction". It turns out, however, that the answer can again be expressed in the change of f due to a infinitesimal change in the x- and y-directions separately.
 
  • #3
HallsofIvy
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If [itex]f(x)= x^2[/itex] and x is a function of t then the derivative of f with respect to t is, by the chain rule, 2x (dx/dt).

Of course, if x is NOT a function of t, then df(x)/dt= (df/dx)(dx/dt)= (df/dx)(0)= 0.
 
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