1
$\begingroup$

I had a little argument with a friend about this. Let $f$ be a differentiable function such that $$g(x,y,z) =xy \ f \left( \frac{y}{x} \right) -z $$

Then, is it mathematically correct to write (I think this is completely ok) $$\nabla g = \left[ y \ f \left( \frac{y}{x} \right) + xy \ f' \left( \frac{y}{x} \right) \left( - \frac{y}{x^2} \right) \right] \hat i + [ \cdots ] \ \hat j - \hat k \tag{1}$$

My friend claims that since $f$ depends on two variables, I cannot just simply write $f'$, and she says that I should have defined $u = \frac{y}{x}$ so that $f'(u)$ would be meaningful.

i.e. she says that the only way to write this gradient is $$\nabla g = \left[ y \ f (u) + xy \ f' (u) \frac{\partial u}{\partial x} \right] \hat i + [ \cdots ] \ \hat j - \hat k \tag{2}$$

I know the way she does in (2) is also true, but does my notation (1) have a problem?

$\endgroup$
2
  • $\begingroup$ I removed the scaled parentheses from the title since the page looks cluttered when loading with this in the title. $\endgroup$
    – AlexR
    Mar 7, 2015 at 14:52
  • 1
    $\begingroup$ She is wrong in saying that you can't write $f'$ and in saying that you 'should define $u$'. You can easily do without defining $u$. Also tell her to replace $f$ by $\sin$. Really, you can't write $\sin'$? She's mistaking $f\circ u$ with $f$. $\endgroup$
    – Git Gud
    Mar 7, 2015 at 15:08

1 Answer 1

1
$\begingroup$

As far as I can understand, you are both right. It is clear that here $f \colon \mathbb{R} \to \mathbb{R}$ and as such $f'$ is perfectly legitimate. You consider the map (with an appropriate domain of definition) $$ \tilde{f}\colon (x,y) \mapsto f(y/x), $$ and this is a function of two variables. But $$ \partial_1\tilde{f}(x,y)=xf'(y/x) $$ is a good piece of notation. Your friend is essentially factorizing $\tilde{f}$ as $$ (x,y) \mapsto y/x=u \mapsto f(y/x), $$ but I can't see any real difference between your approaches.

$\endgroup$
0

You must log in to answer this question.

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