Question

Compute fxy (0, 0) and fyx (0, 0) for the function
f(x,y)={(xy^3)/x+(y)^2, (x,y) != (0,0)
={0, (x,y)=(0,0)
Also, discuss the continuity of fxy and fyx at the origin

Answer 1

fxy and fyx are continuous at the origin.

Explanation:

To find fxy (0, 0), we differentiate f(x, y) with respect to y and then evaluate it at (0, 0). Let's calculate it step by step:

Differentiating f(x, y) with respect to y:

fxy (x, y)

= $\frac{d}{dy} [\frac{xy^{3} }{x} +y^{2}]$

= $\frac{d}{dy} [\frac{xy^{3} }{x}]+\frac{d}{dy}[y^{2}]$

= $\frac{3xy^{2} }{x} +2y$

= $3y^{2} +2y$

Evaluating fxy (0, 0):

fxy (0, 0) = $3(0)^2 + 2(0)$

= 0

Similarly, to find fyx (0, 0), we differentiate f(x, y) with respect to x and then evaluate it at (0, 0):

Differentiating f(x, y) with respect to x:

fyx (x, y)

= $\frac{d}{dx} [\frac{xy^{3} }{x} +y^{2}]$

$\frac{d}{dx} [\frac{xy^{3} }{x}]+\frac{d}{dx}[y^{2}]$

$= (0) + 0$

= 0

Evaluating fyx (0, 0):

fyx (0, 0) = 0

Now, let's discuss the continuity of fxy and fyx at the origin (0, 0).

The functions fxy and fyx are continuous at a point (a, b) if the partial derivatives are continuous in a neighborhood of that point. In this case, both fxy and fyx are constant functions, which means they are continuous at all points, including the origin (0, 0).

Since the partial derivatives are continuous, we can conclude that fxy and fyx are continuous at the origin.

For more such questions on continuous functions:

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