Partial Derivatives

Question

Consider f:R^2→R defined as f(x,y)=(x·y·sin(y))/x^2+y^2 if (x,y) not= 0 and f(x,y)= 0 if (x,y) = (0,0). Is the function continuous at (0,0)? Do the partial derivatives ∂f/∂x and ∂f/∂y exist at (0, 0)? 

Answer 1

Hey Helen,

What makes a function discontinuous? Usually this occurs when the output is not defined for a given input. In this case what values of x,y make the output undefined? This should help you with the continuous part of the question.

As for the partial derivatives lets break them up. The partial derivative with respect to x means that the only variable that can change is x. So it may help to rewrite the equation with a new variable that is more commonly used for a constant value. Let's let y=c and rewrite the problem:

f(x,c) = (xc sin(c))/ (x²+c²) = (c sin(c)) x / (x²+c²)

Hopefully this will make it easier to take the derivative with respect to x while treating c as a constant:

∂f/∂x = c sin(c)( (x²+c²) - x(2x)) / (x²+c²)²

Then put the y back in:

∂f/∂x = y sin(y)( (x²+y²) - 2x²) / (x²+y²)²

The other partial derivative follows in the same manner. Think about whether these derivatives exist at (0,0) the same way you did for the question of continuity in the first part of the problem. Hope this helps!

-Nathan