# Derivative of $\ f (y/x)$

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?

• I removed the scaled parentheses from the title since the page looks cluttered when loading with this in the title. Mar 7, 2015 at 14:52
• 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 $f$. Mar 7, 2015 at 15:08

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.