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Partial derivatives, why is the fy this?

  1. Oct 14, 2005 #1
    Hello everyone,i had a question..i have the following problem, i'm suppose to find the first partial derivatives:
    f(x,y,z,t) = xyz^2*tan(yt);
    I got all the partial derivaties right but the fy, they get:
    fy = xyz^2*sec^2(yt);
    when i do it, i get:
    fy = xz^2*sec^2(yt)*t = txz^2*sec^2(yt);
    Arn't u suppose to think x,z and t are all constants? if thats the case, when u use the chain rule inside the sec^2(yt), d/dt(yt) = t isn't it?
     
  2. jcsd
  3. Oct 14, 2005 #2

    TD

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    Homework Helper

    Hmm, I actually don't agree with either of those answers.

    If all the variables are independant, you have to consider x,z and t as constants when deriving with respect to y, that's correct. But you have to use the product rule as well, there's a y in "xyz^2" and again one in the tangent...
     
  4. Oct 14, 2005 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    If f really is as you write it, that's clearly wrong.
    But that's also clearly wrong!
    Yes, doing a partial derivative, you treat all other variables as constants. But in addition to using the chain rule, you also have to use the product rule!

    fy= xz2tan(yt)+ txyz2sec2(yt)
     
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