Taking the partial derivative of $z=F(x/y)$

Question

I have just started on chapter for partial derivative and I have a very basic question: Please go easy on me since I just started on the topic today.

Let the function be:

$$z=F\left(\frac{x}{y}\right)$$

I am wondering if the reasoning of mine is correct.

$$\frac{\partial z}{\partial x}=\frac{F'\left(\frac{x}{y}\right)}{y}$$ now what I am intrested to know is the logic behind it.

I use the chain rule:

$$F'\left(\frac{x}{y}\right)* ?$$ here is the question: Do I leave $$\frac{1}{y}$$ and differentiate x which is equal to 1.

Second part which obviously is just as easy when you get the hang of it:

$$\frac{\partial z}{\partial y}=-\frac{x F'\left(\frac{x}{y}\right)}{y^2}$$

I use the same logic:

Chain rule:

$$F'\left(\frac{x}{y}\right)* ?$$

now I leave $$\frac{x}{1}$$

And differentiate $$\frac{1}{y}$$

$$F'\left(\frac{x}{y}\right)*x*\frac{-1}{y^2}$$

Is my logic correct or is it something I am missing?

Answer 1

Your answers are fine. The chain rule gives us $$\frac{\partial}{\partial x} f(g(x,y)) = f'(g(x,y))\cdot \frac{\partial g}{\partial x} $$The only real work is the second piece. In calculating $\frac{\partial g}{\partial x}$ we treat anything which does not depend on $x$ as a constant. So $$\frac{\partial }{\partial x} \frac{x}{y} = \frac{1}{y}$$

In the same way that

$$\frac{d }{d x} \frac{x}{2} = \frac{1}{2}$$

Answer 2

You can think about it as ordinary derivatives but with only one of the variables as variable and the others as constants. Your examples: $$\frac{\text{d} F\left(\frac{x}{y_c}\right)}{\text{d} x}=\frac{F'\left(\frac{x}{y_c}\right)}{y_c}$$ $$\frac{\text{d} F\left(\frac{x_c}{y}\right)}{\text{d} y}=x_c\cdot\frac{F'\left(\frac{x_c}{y}\right)}{y^2}$$