Need help with partial derivative: product rule

Question

Given $f(x,y) = \dfrac{(x+1)(y+1)(x+y)}{x^2 y^2}$. Can someone help me with evaluating the critical points?

I did $\dfrac{\partial f}{\partial x}$ and got $-x^2y^4 - x^2 y^2 - 2y^2 x^3 -2xy^4 -2xy^3$ but that's not the correct answer.

The answer should be: $\dfrac{\partial f}{\partial x} = -\dfrac{(y+1)(xy+x+2y)}{x^3 y^2}$.

Should I use the product rule or quotient-rule?

Edit: Using the quotientrule and productrule I get the following: \begin{align*} \dfrac{\partial f}{\partial x} &= (x^2 y^2) \Big[(x+1)(y+1)(x+y)\Big]' - \Big[(x+1)(y+1)(x+y)\Big] 2y^2 x \\ &= x^2 y^2 \Big[\Big((x+1)(y+1)\Big)' (x+y) + (x+1)(y+1) \Big] - \Big[(x+1)(y+1)(x+y) \Big] 2y^2 x \\ &= x^2 y^2 \Big[(y+1)(x+y)+(x+1)(y+1) \Big] - \Big[(x+1)(y+1)(x+y) \Big] 2y^2 x \\ &= x^2 y^2 \Big[xy + y^2 +x +y + xy +x + y + 1 \Big] - \Big[(xy+x+y+1)(x+y) \Big] 2y^2 x \\ &= x^3 y^3 + x^2 y^4 + x^3 y^2 + x^2 y^3 + x^3 y^3 + x^3 y^2 + x^2 y^3 + x^2 y^2 - \Big[ x^2 y + x^2 + xy + x + xy^2 + xy + y^2 + y \Big] 2y^2 x \\ & = -x^2 y^4 - x^2 y^2 - 2x^2 y^3 - 2xy^4 - 2xy^3 \end{align*} But that still not correct. Please, where did I go wrong?

Answer 1

Before differentiating, you can take $log_e$ both sides and that will make the calculations easier.

$$f(x,y)=\frac{(x+1)(y+1)(x+y)}{{x^2}{y^2}}$$ $$log_ef(x,y)=log_e\frac{(x+1)(y+1)(x+y)}{{x^2}{y^2}}$$ $$log_ef(x,y)=log_e(x+1)+log_e(y+1)+log_e(x+y)-2log_ex-2log_ey$$ Implicit differentiate both sides and treat $y$ as a constant, you will get: $$\frac{1}{f(x,y)}\frac{\partial f}{\partial x}=\frac{1}{x+1}+\frac{1}{x+y}-\frac{2}{x}$$ $$\therefore\frac{\partial f}{\partial x}=f(x,y)\left[\frac{1}{x+1}+\frac{1}{x+y}-\frac{2}{x}\right]$$ $$\therefore\frac{\partial f}{\partial x}=\frac{(x+1)(y+1)(x+y)}{{x^2}{y^2}}\left[\frac{1}{x+1}+\frac{1}{x+y}-\frac{2}{x}\right]$$ After simplification, you will get the desired result. However, you need to take note that taking $log_e$ on both sides is only for computation purposes. In theory, one can encounter trouble at those points that the function value is non-positive, which will result the $log_e$ taking operation to be illegal (We are in the real setting). Hope it helps.:)

Answer 2

The function $f$ is a quotient whose numerator and denominator are products. So you have to apply the quotient rule in order to differentiate $f$, but you will also need the product rule once you differentiate the numerator and the denominator. I suggest that you try and do this now, and post your results.

Answer 3

Here's one way that might make this easier. Substitute $u = \frac{1}{x}$ and $v = \frac{1}{y}$ we get

$$f(u, v) = (1+u)(1+v)(u+v) $$

The derivatives can be found using the chain rule

$$ \begin{align} \frac{\partial f}{\partial x} &= \frac{\partial u}{\partial x} \frac{\partial f}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial f}{\partial v} \\&= -\frac{1}{x^2} \, \frac{\partial}{\partial x} \Big[ (1+v)(u^2 + uv + u + v) \Big] \\&= - \frac{1}{x^2} (1+v)(2u + v + 1) \\&= -\frac{1}{x^2} \left(1 + \frac{1}{y}\right)\left(\frac{2}{x} + \frac{1}{y} + 1 \right) \end{align}$$