- #1
- 880
- 938
Homework Statement
Find [itex]\frac{\partial}{\partial x} [/itex] if:
[tex]f(x,y) = \begin{cases}x^2\frac{\sin y}{y}, & y\neq 0\\0, &y=0 \end{cases}[/tex]
Homework Equations
The Attempt at a Solution
If [itex]y\neq 0 [/itex], 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?
[tex]f_x = \frac{\sin y}{y}\lim_{t\to 0}\frac{(x+t)^2 - x^2}{t} = 2x\frac{\sin y}{y}[/tex]. If y = 0 can I say that [itex]\frac{\sin y}{y} = 1[/itex] 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: