# Thread: Partial derivatives with logarithms

1. ## Partial derivatives with logarithms

Ok, so my original function is $\displaystyle f(x,y)=ln(x^2+y^2+1)$ and the problem states to take the partial derivative, first with respect to x, and then with respect to y, and find values of x and y such that both partial derivatives equal zero simultaneously. I understand that with a function such as $\displaystyle x^2+2x+1$ for instance, you can just do substitution with another function of the same type to cancel out one variable to find either x or y then plug it in to find the other variable...however, with this problem, my answers are as follows:

partial derivative with respect to x is $\displaystyle 2x / x^2+y^2+1$ and partial derivative with respect to y is $\displaystyle 2y / x^2+y^2+1$.

How would you guys recommend I solve this? Set the derivatives equal to each other and solve for a variable or something else?

2. We have:

$\displaystyle z = f(x,y) = ln(x^2 + y^2 + 1)$

You also provided:

$\displaystyle \frac{\partial z}{\partial x} = \frac{2x}{x^2 + y^2 + 1}$

$\displaystyle \frac{\partial z}{\partial x} = \frac{2y}{x^2 + y^2 + 1}$

If they are to equal zero simultaneously, then it is a requirement that:

$\displaystyle \frac{\partial z}{\partial x} = \frac{\partial z}{\partial y}$

Let's pretend for a second that these were variables we could manipulate:

$\displaystyle \partial z = \partial x\frac{\partial z}{\partial y}$

$\displaystyle \frac{\partial z}{\partial z} = \frac{\partial x}{\partial y}$

$\displaystyle 1 = \frac{\partial x}{\partial y}$

$\displaystyle 1 = \frac{x}{y}$

$\displaystyle y = x$

If you did the math with each individual equation, you would find that this is true (Keep in mind, you will not be able to treat the partials like this every time). Now, if x equals y, the only way for them to equal zero simultaneously in this equation would be $\displaystyle x = 0$ and $\displaystyle y = 0$.

There you go.