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

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


    User Avatar
    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


    User Avatar
    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)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook