Discuss the continuity and differentiability of the function f(x,y)=x2yx2+y2 at the point (0,0)

Given that f(x,y)=x2yx2+y2 To find the continuity, lim(x,y)→(0,0)f(x,y)=lim(x,y)→(0,0)x2yx2+y2 =lim(x,y)→(0,0)x3x2+x2(y=x) =lim(x,y)→(0,0)x2 =0 Hence f(x,y)=x2yx2+y2 is continuous at (0,0) Now to find the derivative for f with respect to x, fx=2xyx2+y2−2xyx2+y2[1−x2x2+y2] =2xyx2+y2y2x2+y2 =2xy3x2+y2 Take the derivative for f with respect to y, fy=x2x2+y2−x2y×2y(x2+y2)2 =x2x2+y2[1−2y2x2+y2] =x2(x2−y2)(x2+y2)2 The differentiability for f at (0,0) is Df=(fx,fy)=(0,0).

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nitraiddQ

Answered question

2021-10-29

Discuss the continuity and differentiability of the function

f (x, y) = \frac{x^{2} y}{x^{2} + y^{2}}

at the point (0,0)

Answer & Explanation

un4t5o4v

Skilled2021-10-30Added 105 answers

Given that

f (x, y) = \frac{x^{2} y}{x^{2} + y^{2}}

To find the continuity,

lim_{(x, y) \to (0, 0)} f (x, y) = lim_{(x, y) \to (0, 0)} \frac{x^{2} y}{x^{2} + y^{2}}

= lim_{(x, y) \to (0, 0)} \frac{x^{3}}{x^{2} + x^{2}} (y = x)

= lim_{(x, y) \to (0, 0)} \frac{x}{2}

= 0

Hence

f (x, y) = \frac{x^{2} y}{x^{2} + y^{2}}

is continuous at (0,0)
Now to find the derivative for f with respect to x,

f_{x} = \frac{2 x y}{x^{2} + y^{2}} - \frac{2 x y}{x^{2} + y^{2}} [1 - \frac{x^{2}}{x^{2} + y^{2}}]

= \frac{2 x y}{x^{2} + y^{2}} \frac{y^{2}}{x^{2} + y^{2}}

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

Take the derivative for f with respect to y,

f_{y} = \frac{x^{2}}{x^{2} + y^{2}} - \frac{x^{2} y \times 2 y}{{(x^{2} + y^{2})}^{2}}

= \frac{x^{2}}{x^{2} + y^{2}} [1 - \frac{2 y^{2}}{x^{2} + y^{2}}]

= \frac{x^{2} (x^{2} - y^{2})}{{(x^{2} + y^{2})}^{2}}

The differentiability for f at (0,0) is

D f = (f_{x}, f_{y}) = (0, 0)

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