The function $f: \mathbb{R} \longrightarrow \mathbb{R}$ is defined by the rule
$$f(x,y) = \begin{cases} \frac{x^5y}{x^4+y^2}, & (x,y) \neq (0,0), \\ 0, & (x,y)=(0,0). \end{cases}$$
Evaluate $f_x(0,y)$, $f_x(0,0)$, $f_y(x,0)$, and $f_y(0,0)$.
(The definition of partial derivative as a limit is the recommended method.)
My question is really a conceptual question, that I am struggling to understand. At the point $(0,0)$, the function is defined to be $0$. So, finding the partial derivatives at those points, why would you not use $f(x,y)=0$?
My instincts tell me that no, you would use the other part of the function (partly because the question suggests using the limit definition): $\dfrac{x^5y}{x^4+y^2}$, $(x,y) \neq (0,0)$. But why is this?
Secondly, in general, why would you use the limit definition of a derivative, as opposed to simply taking the derivative mechanically? What is the difference?
Thanks for any help!
Best Answer
First: Your instinct is correct. Derivatives are "local information." This means that the derivative of a function at any point is computed by using information about the function's nearby values. For example, consider the limit definition of a derivative of a single-variable function: $$f'(x) = \lim_{h\to x}\frac{f(h) - f(x)}{h - x}.$$ In order to compute this, we need to understand all of the values $f$ takes at the $h$ values near $x$.
Second: Taking the derivative "mechanically" is just cranking through a set of rules that you have learned for some particularly nice functions (products, sums, quotients of differentiable functions; exponentials, trig functions, polynomials, etc). You can't take the derivative at $0$ mechanically here because, at zero, your function is not one of those particularly nice functions.
Since you can't use the differentiation rules, you're stuck with computing the derivative straight from the limit definition.