Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivitive of X with Respect to Time

  1. Feb 13, 2010 #1
    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
  2. jcsd
  3. Feb 13, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    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.
  4. Feb 13, 2010 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    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.
    Last edited: Feb 14, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook