Finding partial derivatives of a given function

Question

I must calculate $\displaystyle \frac{\partial{f}}{\partial{x}}$ and $\displaystyle\frac{\partial{f}}{\partial{y}}$ from $f(x,y)=\displaystyle\frac{4 \cdot \pi^2 \cdot x}{y^2}$. $$\displaystyle \frac{\partial{f}}{\partial{x}}=\frac{4 \cdot \pi^2}{y^2}$$ $$\displaystyle \frac{\partial{f}}{\partial{y}}=\frac{-8 \cdot \pi^2 \cdot x}{y^3}$$ Is it correct? Thanks in advance!

Answer 1

As it is a homework, let us use the definitions of partial derivatives to check results, which are normally quickly obtained by following the rule "keep all fixed except for the variable w.r.t. you compute the partial derivative and use results from Analysis I/Calculus".

Then

$$\frac{\partial f}{\partial x}(x,y):=\lim_{h\rightarrow 0}\frac{f(x+h,y)-f(x,y)}{h}= \lim_{h\rightarrow 0}\frac{4\pi^2}{y^2h}(x+h-x)= \frac{4\pi^2}{y^2}.$$

$$\frac{\partial f}{\partial y}(x,y):=\lim_{h\rightarrow 0}\frac{f(x,y+h)-f(x,y)}{h}= \lim_{h\rightarrow 0}\frac{4\pi^2x}{h}\left(\frac{1}{(y+h)^2}-\frac{1}{y^2}\right)= \lim_{h\rightarrow 0}\frac{4\pi^2x}{h}\left(\frac{h(-2y-h)}{y^2(y+h)^2}\right)= \lim_{h\rightarrow 0}\frac{4\pi^2x(-2y-h)}{y^2(y+h)^2}=\frac{-8\pi^2x}{y^3}.$$

The computations in the OP are correct.