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Implicit Differentiation z=f(x/y) meaning

  1. Jan 2, 2016 #1
    Mod note: Moved from the Homework section
    1. The problem statement, all variables and given/known data

    This might seem like a stupid question but I'm unsure what z= ƒ(x/y) means? I'm not sure how I would find ∂z/∂x , ∂z/∂y just from this statement either.

    Thank you
    2. Relevant equations


    3. The attempt at a solution
     
    Last edited by a moderator: Jan 2, 2016
  2. jcsd
  3. Jan 2, 2016 #2

    PeroK

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    You could try taking ##f## to be some simple function. For example ##f(X) = X^2##, where I've used ##X## to define the function to avoid confusion with ##x, y##.
     
  4. Jan 2, 2016 #3

    Mark44

    Staff: Mentor

    It means that z is a function of the quotient x/y. For example, ##z = (x/y)^2 + 3(x/y)##

    To find the partial derivatives, you'll need to use the chain rule.
     
  5. Jan 2, 2016 #4
    I'm sorry, I'm just really unsure how to apply the chain rule. So is ∂z/∂t = ∂/∂x(x/y) * F' ? But I'm unsure how to find all the other ones. I have the solutions attached to this post but I have no idea how to get them.
    Thank You
     

    Attached Files:

  6. Jan 2, 2016 #5

    vela

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    What are ##t## and ##F##?

    Can you calculate ##\frac{\partial z}{\partial x}## if ##z=f(u(x,y))## where ##u## is a function of ##x## and ##y##?
     
  7. Jan 2, 2016 #6
    So i would let u=x/y so then

    $$ \frac{\partial z}{\partial x} = \frac{\partial z }{\partial u} \frac{\partial u }{\partial x} = \frac{f'}{y} $$

    cause I'm guessing $$\frac{\partial z }{\partial u} = f'$$ or am i completely wrong?
     
  8. Jan 2, 2016 #7

    vela

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    That's right.
     
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