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Partial Derivate of multivariable cos

  1. Aug 21, 2011 #1
    1. The problem statement, all variables and given/known data

    Partial derivative of R = vt cos x with respect to t

    2. Relevant equations


    3. The attempt at a solution

    Keeping the other variables constant, I get 'v cos x'. (t is 1)

    I just want to check this, because it's part of a larger problem and I feel my answer is way off because of this function. I'm a bit rusty on my calculus, it's been ... 13 years since I did it in highschool!

    Also, if I calculate the partial derivative with respect to x, I get '-vt sin x'?
  2. jcsd
  3. Aug 21, 2011 #2


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    If v and x are functions of t, then vt cos(x) is actually a product of three terms. You will need to differentiate this using the Product Rule.
  4. Aug 21, 2011 #3
    OK, I think I'm confused as to what a Partial Derivative is then. According to the wiki page example:



    In the above example the xy term becomes y, whereas if the product rule was used, it would be x'y + y'x = y + x, or have I missed something?

    Let me elaborate on the problem I'm trying to solve, it's an error bound and the equation is:

    |Error(R)| <= |dR/dv(v)||Error(v)| + |dR/dt(t)||Error(t)| + |dR/dx(x)||Error(x)|

    where dR/dv(v) is the Partial Derivative of R = vt cos x wrt v (multiplied by the error in v), and similarly for the two other derivatives wrt t and x.
    Last edited by a moderator: Apr 26, 2017
  5. Aug 21, 2011 #4


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    I take it x is the angle a projectile is being launched at, so it's not a function of time. Usually, the angle is denoted by a Greek letter, like θ, and the displacement in the horizontal direction, which is a function of time, is denoted by x.

    So as far as your original post goes, your partial derivatives are correct.
  6. Aug 21, 2011 #5
    Oh, yes it is an angle (in radians) and the symbol was phi, but I didn't have it at hand, so I just used x. I didn't realise that would alter the equation. Thanks for the clarity! :)
  7. Aug 21, 2011 #6


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    You have missed that the derivative of y with respect to x is 0. So one of the terms you get when you use the product rule on "xy" is 0. By the way, you shouldn't write (xy)'=x'y+xy', because this notation doesn't reveal what functions you're taking the derivative of. I like the notation [tex]\frac{d}{dx}(xy)=\bigg(\frac{d}{dx}x\bigg)y+x\bigg(\frac{d}{dx}y\bigg)=1y+x0=y.[/tex] Alternatively, you can define two functions f and g by [itex]f(x)=x[/itex] for all x, and [itex]g(x)=y[/itex] for all x. (Note that this makes g a constant function). Then you can write (fg)'(x)=f'(x)g(x)+f(x)g'(x)=1y+x0=y.

    Regarding what a partial derivative is...don't think of it as something different from an ordinary derivative. It isn't. If you're asked to compute the partial derivative of xy2 with respect to x, it means this: Let f be the function defined by f(t)=ty2 for all t. Find f'(x) (i.e. the derivative of f, evaluated at x). If you're asked to compute the partial derivative of xy2 with respect to y, it means this: Let g be the function defined by g(t)=xt2 for all t. Find g'(y) (i.e. the derivative of g, evaluated at y).

    The partial derivative of vt cos x with respect to t...that's the ordinary derivative of the function that takes t to vt cos x (as opposed to e.g. the function that takes x to vt cos x).

    See what I mean when I say that a partial derivative isn't really something different from an ordinary derivative? The "with respect to" part of it is just telling you which function to take an ordinary derivative of.
    Last edited by a moderator: Apr 26, 2017
  8. Aug 21, 2011 #7
    Thanks Fredrik, that clears things up a lot, I appreciate the thorough explanation.
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