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Partial derivatives

  1. Mar 29, 2004 #1


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    ok i need help with a few questions. i'll post the question first and then what i get as an answer, the first one is the partial derivative with respect to x and the second one with respect to y. these are the even number questions from my textbook and they dont have answers to them so if someone could check them, i would really appreciate it; thanks.

    a) z = yln|x|
    b) z = x^y
    c) z = xy^2/(x^2+y^2)
    d) z = arctan(x/y)

    a) 1/x, 1
    b) yx^(y-1), (x^y)(ln|x|)
    c) [(x^2)(x^2+y^2)-xy^2(2s)]/(x^2+y^2)^2,
    d) i dont even know how to do this one.
  2. jcsd
  3. Mar 29, 2004 #2
    Be sure to treat the part of the expression that doesn't have the variable you're differentiating with respect to as a constant.

    For the arctan problems, take the tangent of both sides and differentiate implicitly.

  4. Mar 29, 2004 #3


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    um, do you think you could explain that? thanks. i've read over my profs notes and they make sense but they are the simple examples.
  5. Mar 29, 2004 #4
    Okay. In the first problem,

    z = y ln|x|

    If we're differentiating with respect to x, we're going to treat anything with a y in it as constant. That is, we're going to treat y multiplier in front as a constant. So essentially we're differentiating

    z = c*ln|x|

    where c is some constant. I'm sure you know how to differentiate that to get

    dz/dx = c/x = y/x (since c = y).

    Now to differentiate with respect to y, we treat anything with an x as a constant. That would be the expression ln|x|. So we're differentiating

    z = y*c

    which is

    dz/dy = c = ln|x| (since c = ln|x|)

    For the last one, you need to take the tangent of each side to get the equation

    tan(z) = x/y

    We then use the chain rule to differentiate tan(z), i.e.

    [tex]\frac{\partial f(z)}{\partial x} = \frac{\partial f(z)}{\partial z}\frac{\partial z}{\partial x}[/tex]

    it is the same for differentiating with respect to y, just replace the x's in the above equation with y's.

    That gives us the left-hand side. The right-hand side you differentiate normally. Once you have the left- and right-hand sides, solve for dz/dx (or dz/dy, if you're differentiating with respect to y) and that will be your answer. If you have an expression with z in your answer, substitute it away using the problem statement.

  6. Mar 30, 2004 #5

    If we have d/dx [tan x]...doesn't that equal 1/(1+x^2) ??? you should know that from calc1 or 2 right? So just apply the same form to tan x/y, holding y constant for (partial x) and x costant for (partial y). I think this would be right:

    (partial x) = (1/y)/(1+(x/y)^2)


    (partial y) = (-x/y^2)/(1+(x/y)^2)

    Check my work though, not totally sure.
  7. Mar 30, 2004 #6


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    "(partial x) = (1/y)/(1+(x/y)^2)"

    Yes, although it might be a good idea to multiply both numerator and denominator by y2 to get [itex]\frac{\partial f}{\partial x}= \frac{y}{x^2+y^2}[/itex]

    "(partial y) = (-x/y^2)/(1+(x/y)^2)"

    Excellent! Now multiply both numerator and denominator by y2 to get
    [itex]\frac{\partial f}{\partial y}= \frac{-x}{x^2+ y^2}[/itex]
    Last edited: Mar 30, 2004
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