# Partial differentiation

1. Mar 12, 2015

### nuuskur

1. The problem statement, all variables and given/known data
Find $\frac{\partial}{\partial x}$ if:
$$f(x,y) = \begin{cases}x^2\frac{\sin y}{y}, & y\neq 0\\0, &y=0 \end{cases}$$

2. Relevant equations

3. The attempt at a solution
If $y\neq 0$, then it's simple, but I get confused about the second part. How can I exactly utilize the limit definition of a derivative when I'm told that y=0?
$$f_x = \frac{\sin y}{y}\lim_{t\to 0}\frac{(x+t)^2 - x^2}{t} = 2x\frac{\sin y}{y}$$. If y = 0 can I say that $\frac{\sin y}{y} = 1$ since that would be the limit as y approaches 0, but y is already fixed at 0. Does the partial derivative not exist if y = 0?

Last edited: Mar 12, 2015
2. Mar 12, 2015

### SteamKing

Staff Emeritus
Wow! You weren't kidding when you said you're confused.

Since you want to find ∂f(x,y) / ∂x, then y is treated as a constant, so it is irrelevant what the limit of sin(y)/y is.

3. Mar 12, 2015

### nuuskur

And I am treating it as a constant. When the arguments are taken only on the x-axis I would say that the derivative is $\frac{0}{0}2x$, which is gibberish and leads me to believe that the function is not partially differentiable with respect to x when y=0. Is this conclusion correct, incorrect and/or incomplete?

4. Mar 12, 2015

### SteamKing

Staff Emeritus
Since f(x,y) = 0 when y = 0, then what is ∂ 0 / ∂x ? Remember, f(x,y) = 0 when y = 0 presumably holds for any and all values of x.

5. Mar 12, 2015

### nuuskur

Oh right, I am dumb. It's 0.