Differentiability for a multivariable piecewise function
$begingroup$
$$f(x,y)=
begin{cases}
dfrac{sinh(xy)+2x^2-y}{1+x^2} &&,text{if } |y|leqsqrt{|x|}text{ and } y=-x^2\
dfrac{x^2+(2-alpha)y+y^2}{y+x^2} &&,text{otherwise}
end{cases}
$$
I have to study the differentiability of $f$ in the origin, and say if $f$ is differentiable for at least one real $alpha$.
When it comes to study the differentiability of the function I'm stuck for what piece of function I have to choose, my first thought was to study both together but I'm not sure.
So that would be:
$$f(x,y)=
begin{cases}
limlimits_{h,kto(0,0)} dfrac{sinh(hk)+2h^2-k-partial_x(0,0)-partial_y(0,0)-f(0,0)}{(1+h^2)sqrt{h^2+k^2}}\
limlimits_{h,kto(0,0)} dfrac{h^2(2-alpha)k+k^2-partial_x(0,0)-partial_y(0,0)-f(0,0)}{(1+h^2)sqrt{h^2+k^2}}
end{cases}
$$
and they have to be the same limit.
Same thing with the partial derivatives:
$$begin{align}
partial_xf(x,y)&=
begin{cases}
limlimits_{hto0} dfrac{2h^2}{h+h^3}=0\
limlimits_{hto0} dfrac{h^2}{h^2}=1
end{cases}\
partial_yf(x,y)&=
begin{cases}
limlimits_{kto0} dfrac{-k}{k}=-1\
limlimits_{kto0} dfrac{(2-alpha)k+k^2}{k^2}=+infty
end{cases}
end{align}$$
So $f$ is not differentiable for any $alpha$
Is this correct?
multivariable-calculus
edited Dec 20 '18 at 5:16
Dylan
12.5k31026
asked Dec 18 '18 at 21:15
ArchimedessArchimedess
235
$endgroup$
add a comment |
$begingroup$
$$f(x,y)=
begin{cases}
dfrac{sinh(xy)+2x^2-y}{1+x^2} &&,text{if } |y|leqsqrt{|x|}text{ and } y=-x^2\
dfrac{x^2+(2-alpha)y+y^2}{y+x^2} &&,text{otherwise}
end{cases}
$$
I have to study the differentiability of $f$ in the origin, and say if $f$ is differentiable for at least one real $alpha$.
When it comes to study the differentiability of the function I'm stuck for what piece of function I have to choose, my first thought was to study both together but I'm not sure.
So that would be:
$$f(x,y)=
begin{cases}
limlimits_{h,kto(0,0)} dfrac{sinh(hk)+2h^2-k-partial_x(0,0)-partial_y(0,0)-f(0,0)}{(1+h^2)sqrt{h^2+k^2}}\
limlimits_{h,kto(0,0)} dfrac{h^2(2-alpha)k+k^2-partial_x(0,0)-partial_y(0,0)-f(0,0)}{(1+h^2)sqrt{h^2+k^2}}
end{cases}
$$
and they have to be the same limit.
Same thing with the partial derivatives:
$$begin{align}
partial_xf(x,y)&=
begin{cases}
limlimits_{hto0} dfrac{2h^2}{h+h^3}=0\
limlimits_{hto0} dfrac{h^2}{h^2}=1
end{cases}\
partial_yf(x,y)&=
begin{cases}
limlimits_{kto0} dfrac{-k}{k}=-1\
limlimits_{kto0} dfrac{(2-alpha)k+k^2}{k^2}=+infty
end{cases}
end{align}$$
So $f$ is not differentiable for any $alpha$
Is this correct?
multivariable-calculus
edited Dec 20 '18 at 5:16
Dylan
12.5k31026
asked Dec 18 '18 at 21:15
ArchimedessArchimedess
235
$endgroup$
add a comment |
$begingroup$
$$f(x,y)=
begin{cases}
dfrac{sinh(xy)+2x^2-y}{1+x^2} &&,text{if } |y|leqsqrt{|x|}text{ and } y=-x^2\
dfrac{x^2+(2-alpha)y+y^2}{y+x^2} &&,text{otherwise}
end{cases}
$$
I have to study the differentiability of $f$ in the origin, and say if $f$ is differentiable for at least one real $alpha$.
When it comes to study the differentiability of the function I'm stuck for what piece of function I have to choose, my first thought was to study both together but I'm not sure.
So that would be:
$$f(x,y)=
begin{cases}
limlimits_{h,kto(0,0)} dfrac{sinh(hk)+2h^2-k-partial_x(0,0)-partial_y(0,0)-f(0,0)}{(1+h^2)sqrt{h^2+k^2}}\
limlimits_{h,kto(0,0)} dfrac{h^2(2-alpha)k+k^2-partial_x(0,0)-partial_y(0,0)-f(0,0)}{(1+h^2)sqrt{h^2+k^2}}
end{cases}
$$
and they have to be the same limit.
Same thing with the partial derivatives:
$$begin{align}
partial_xf(x,y)&=
begin{cases}
limlimits_{hto0} dfrac{2h^2}{h+h^3}=0\
limlimits_{hto0} dfrac{h^2}{h^2}=1
end{cases}\
partial_yf(x,y)&=
begin{cases}
limlimits_{kto0} dfrac{-k}{k}=-1\
limlimits_{kto0} dfrac{(2-alpha)k+k^2}{k^2}=+infty
end{cases}
end{align}$$
So $f$ is not differentiable for any $alpha$
Is this correct?
multivariable-calculus
edited Dec 20 '18 at 5:16
Dylan
12.5k31026
asked Dec 18 '18 at 21:15
ArchimedessArchimedess
235
$endgroup$
$$f(x,y)=
begin{cases}
dfrac{sinh(xy)+2x^2-y}{1+x^2} &&,text{if } |y|leqsqrt{|x|}text{ and } y=-x^2\
dfrac{x^2+(2-alpha)y+y^2}{y+x^2} &&,text{otherwise}
end{cases}
$$
I have to study the differentiability of $f$ in the origin, and say if $f$ is differentiable for at least one real $alpha$.
When it comes to study the differentiability of the function I'm stuck for what piece of function I have to choose, my first thought was to study both together but I'm not sure.
So that would be:
$$f(x,y)=
begin{cases}
limlimits_{h,kto(0,0)} dfrac{sinh(hk)+2h^2-k-partial_x(0,0)-partial_y(0,0)-f(0,0)}{(1+h^2)sqrt{h^2+k^2}}\
limlimits_{h,kto(0,0)} dfrac{h^2(2-alpha)k+k^2-partial_x(0,0)-partial_y(0,0)-f(0,0)}{(1+h^2)sqrt{h^2+k^2}}
end{cases}
$$
and they have to be the same limit.
Same thing with the partial derivatives:
$$begin{align}
partial_xf(x,y)&=
begin{cases}
limlimits_{hto0} dfrac{2h^2}{h+h^3}=0\
limlimits_{hto0} dfrac{h^2}{h^2}=1
end{cases}\
partial_yf(x,y)&=
begin{cases}
limlimits_{kto0} dfrac{-k}{k}=-1\
limlimits_{kto0} dfrac{(2-alpha)k+k^2}{k^2}=+infty
end{cases}
end{align}$$
So $f$ is not differentiable for any $alpha$
Is this correct?
multivariable-calculus
multivariable-calculus
edited Dec 20 '18 at 5:16
Dylan
12.5k31026
asked Dec 18 '18 at 21:15
ArchimedessArchimedess
235
edited Dec 20 '18 at 5:16
Dylan
12.5k31026
asked Dec 18 '18 at 21:15
ArchimedessArchimedess
235
edited Dec 20 '18 at 5:16
Dylan
12.5k31026
edited Dec 20 '18 at 5:16
Dylan
12.5k31026
edited Dec 20 '18 at 5:16
Dylan
12.5k31026
12.5k31026
asked Dec 18 '18 at 21:15
ArchimedessArchimedess
235
asked Dec 18 '18 at 21:15
ArchimedessArchimedess
235
asked Dec 18 '18 at 21:15
ArchimedessArchimedess
235
235
add a comment |
add a comment |
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