Question #121883
find fx(0,0) and fx(x,y) where (x,y) not equal to (0,0) for the following function f(x,y) = {xy^3/(x^2 + y^2),(x,y) not equal to (0,0), 0, (x,y)=(0,0)
Is fx continuous at (0,0)?
1
Expert's answer
2020-06-15T19:27:13-0400
f(x,y)={xy3x2+y2if (x,y)(0,0)0if (x,y)=(0,0)f(x,y) = \begin{cases} \dfrac{xy^3 }{x^2 +y^2} &\text{if } (x, y)\not=(0, 0) \\ 0 &\text{if } (x,y)=(0,0) \end{cases}

If (x,y)(0,0)(x, y)\not=(0, 0)


fx=fx=x(xy3x2+y2)=f_{x}=\dfrac{\partial f}{\partial x}=\dfrac{\partial }{\partial x}( \dfrac{xy^3 }{x^2 +y^2} )=

=y3(x2+y2)2x2y3(x2+y2)2=y3(y2x2)(x2+y2)2=\dfrac{y^3(x^2+y^2)-2x^2y^3 }{(x^2 +y^2)^2}=\dfrac{y^3(y^2-x^2)}{(x^2 +y^2)^2}

fx(0,0)=limh0f(0+h,0)f(0,0)h=f_x(0,0)=\lim\limits_{h\to 0}\dfrac{f(0+h,0)-f(0,0) }{h}=

=limh0f(0+h,0)f(0,0)h==\lim\limits_{h\to 0}\dfrac{f(0+h,0)-f(0,0) }{h}=

=limh0h(0)3h2+(0)20h=0=\lim\limits_{h\to 0}\dfrac{ \dfrac{h(0)^3 }{h^2 +(0)^2}-0 }{h}=0


fx(x,y)={y3(x2y2)(x2+y2)2if (x,y)(0,0)0if (x,y)=(0,0)f_x(x,y) = \begin{cases} \dfrac{y^3(x^2-y^2) }{(x^2 +y^2)^2} &\text{if } (x, y)\not=(0, 0) \\ 0 &\text{if } (x,y)=(0,0) \end{cases}

Let x=rcosθ,y=rsinθx=rcos\theta, y=r\sin\theta


lim(x,y)0fx(x,y)=lim(x,y)0y3(x2y2)(x2+y2)2=\lim\limits_{(x, y)\to 0}f_x(x, y)=\lim\limits_{(x, y)\to 0}\dfrac{y^3(x^2-y^2) }{(x^2 +y^2)^2}=

=limr0r3sin3θr2(cos2θsin2θ)r4==\lim\limits_{r\to 0}\dfrac{r^3\sin^3\theta r^2(\cos^2\theta-\sin^2\theta) }{r^4} =

=limr0rsin3θ(cos2θsin2θ)=0=fx(0,0)=\lim\limits_{r\to 0}r\sin^3\theta(\cos^2\theta-\sin^2\theta)=0=f_x(0,0)

The function fx(x,y)f_x(x, y) is continuous at (0,0).(0,0).



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