0
$\begingroup$

I have a simple chain rule homework problem. I feel like im missing something here. The statement of the problem:

Determine $$\frac{\partial z}{\partial x} \text{ when } z = xy f \left( \frac{x}{y} \right).$$

My attempt:

$$\frac{\partial z}{\partial x} = y f \left( \frac{x}{y} \right) + xf' \left( \frac{x}{y} \right).$$

just by product rule and simple chain rule from single variable calculus. I'm kind of weirded out by this, i feel like im overlooking something. The book's notation (Marsden Vector Calculus 5th ed) is much more complicated looking than what I'm doing here.

Am I fudging the bucket? If I am what have I missed conceptually?

$\endgroup$
2
  • 2
    $\begingroup$ It seems all right! $\endgroup$
    – Emo
    Commented Feb 10, 2014 at 20:20
  • $\begingroup$ You did it correctly, since in the second term you get $xyf^{\prime}(x/y)(1/y)=xf^{\prime}(x/y)$. $\endgroup$
    – user84413
    Commented Feb 10, 2014 at 20:26

1 Answer 1

1
$\begingroup$

Your answer appears to be correct. To follow up and make your answer less ambiguous. You should use $\frac{\partial f}{\partial x}(\frac{x}{y})$ because f is a function of more than one variable.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .