What is the partial derivative of f(x,y) with respect to x if y=0?

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The discussion centers on finding the partial derivative of the function f(x,y) with respect to x when y is set to 0. The function is defined as f(x,y) = x²(sin y)/y for y ≠ 0 and f(x,y) = 0 for y = 0. It is clarified that when y = 0, the function simplifies to f(x,y) = 0 for all x, making the partial derivative ∂f/∂x equal to 0. The confusion arises from the limit of sin(y)/y as y approaches 0, but it is established that this limit does not apply when y is fixed at 0. Ultimately, the conclusion is that the partial derivative exists and is equal to 0 when y = 0.
nuuskur
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Homework Statement


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}

Homework Equations

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?
 
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nuuskur said:

Homework Statement


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}

Homework Equations

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?

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.
 
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?
 
nuuskur said:
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 or incorrect?

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.
 
SteamKing said:
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.
Oh right, I am dumb. It's 0.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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