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Partial differentiation

  1. Mar 12, 2015 #1
    1. The problem statement, all variables and given/known data
    Find [itex]\frac{\partial}{\partial x} [/itex] if:
    [tex]f(x,y) = \begin{cases}x^2\frac{\sin y}{y}, & y\neq 0\\0, &y=0 \end{cases}[/tex]

    2. Relevant equations

    3. The attempt at a solution
    If [itex]y\neq 0 [/itex], then it's simple, but I get confused about the second part. How can I exactly utilize the limit definition of a derivative when I'm told that y=0?
    [tex]f_x = \frac{\sin y}{y}\lim_{t\to 0}\frac{(x+t)^2 - x^2}{t} = 2x\frac{\sin y}{y}[/tex]. If y = 0 can I say that [itex]\frac{\sin y}{y} = 1[/itex] since that would be the limit as y approaches 0, but y is already fixed at 0. Does the partial derivative not exist if y = 0?
    Last edited: Mar 12, 2015
  2. jcsd
  3. Mar 12, 2015 #2


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    Wow! You weren't kidding when you said you're confused.

    Since you want to find ∂f(x,y) / ∂x, then y is treated as a constant, so it is irrelevant what the limit of sin(y)/y is.
  4. Mar 12, 2015 #3
    And I am treating it as a constant. When the arguments are taken only on the x-axis I would say that the derivative is [itex]\frac{0}{0}2x [/itex], which is gibberish and leads me to believe that the function is not partially differentiable with respect to x when y=0. Is this conclusion correct, incorrect and/or incomplete?
  5. Mar 12, 2015 #4


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    Since f(x,y) = 0 when y = 0, then what is ∂ 0 / ∂x ? Remember, f(x,y) = 0 when y = 0 presumably holds for any and all values of x.
  6. Mar 12, 2015 #5
    Oh right, I am dumb. It's 0.
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