text: # Evaluate the partial derivatives of the following function:
  
  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!
  
  -
  When you compute the derivative of a single-variable function, it is not just the value of the function at the point that matters, but also the values of the function near but not at the point. You use both when you compute $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$Here you use the limit because "taking the derivative mechanically" requires a single formula for the function in a neighborhood of the point in question, which you don't have here. Think about how you would compute the derivative at $0$ of the function $f(x)$ given by $f(x) = x^2\sin(1/x)$ if $x\neq 0$ and $f(0)=0$. – Arturo Magidin Jun 6 '12 at 4:46
  @ArturoMagidin I have struggled for weeks to grasp this concept, thank you so much for explaining it clearly to me. Actually, the example you gave, is almost the same question I encountered that I struggled to answer a few weeks ago. I couldn't understand why I wouldn't just take the derivative, as opposed to using the limit definition. Now I get it! Thanks! – JackReacher Jun 6 '12 at 5:09
  
  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.
  You can take the derivative, but not with the derivative rules. The problem isn't that the function has different values near $(0,0)$. Just being defined piecewise, like $f$, means that at the points where the definition changes pieces, it's not any of the functions the differentiation rules apply to, so you can't use them. – Neal Jun 6 '12 at 5:33
  @mathstudent You might be skipping some extra algebraic simplification. Looking at it, for $f_y$ and for $f_x$ both at $(0,0)$, I don't see $0/0$. – Neal Jun 6 '12 at 12:38
explanation: The document is a discussion about evaluating the partial derivatives of a piecewise-defined function. It contains a math question and an answer to that question. The question is about why the limit definition of the derivative is used instead of simply using the value of the function at the point of interest, and the answer explains that derivatives are local information and require the use of nearby values of the function, which is why the limit definition is necessary in this case.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Evaluate the partial derivatives of the following function:
  
  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!
  
  -
  When you compute the derivative of a single-variable function, it is not just the value of the function at the point that matters, but also the values of the function near but not at the point. You use both when you compute $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$Here you use the limit because "taking the derivative mechanically" requires a single formula for the function in a neighborhood of the point in question, which you don't have here. Think about how you would compute the derivative at $0$ of the function $f(x)$ given by $f(x) = x^2\sin(1/x)$ if $x\neq 0$ and $f(0)=0$. – Arturo Magidin Jun 6 '12 at 4:46
  @ArturoMagidin I have struggled for weeks to grasp this concept, thank you so much for explaining it clearly to me. Actually, the example you gave, is almost the same question I encountered that I struggled to answer a few weeks ago. I couldn't understand why I wouldn't just take the derivative, as opposed to using the limit definition. Now I get it! Thanks! – JackReacher Jun 6 '12 at 5:09
  
  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.
  You can take the derivative, but not with the derivative rules. The problem isn't that the function has different values near $(0,0)$. Just being defined piecewise, like $f$, means that at the points where the definition changes pieces, it's not any of the functions the differentiation rules apply to, so you can't use them. – Neal Jun 6 '12 at 5:33
  @mathstudent You might be skipping some extra algebraic simplification. Looking at it, for $f_y$ and for $f_x$ both at $(0,0)$, I don't see $0/0$. – Neal Jun 6 '12 at 12:38
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion about evaluating the partial derivatives of a piecewise-defined function. It contains a math question and an answer to that question. The question is about why the limit definition of the derivative is used instead of simply using the value of the function at the point of interest, and the answer explains that derivatives are local information and require the use of nearby values of the function, which is why the limit definition is necessary in this case.
  
  Label: question-and-answer