Show $f(x,y)=x^2\log(x^4+y^2)$ is differentiable at $\vec 0$ I have to show that $f$ is differentiable at $\vec 0$, where
$$
f(x,y)=x^2\log(x^4+y^2),
$$
and $f(0,0)=0$.
I’ve already shown that $f$ is continuous at $\vec 0$. I started off by calculation the first partial derivative:
$$
D_1f(\vec 0)=\lim_{t\to 0}t\log t^4.
$$
However, I don’t know how to calculate this limit even. I looked at the plot, and it seems that $D_f(\vec 0)=D_2f(\vec 0)=0$, so apparently $t$ goes faster to zero then $\log t^4$ goes to minus infinity. How can I show this? Can I use Taylor? Should I evaluate then at $x=1$? This would yield:
$$
\log x=(x-1)-\frac{(x-1)^2}{2}+O((x-1)^3).
$$
Is this the way to go? I've never expanded $\log x$ before like this, and I'm unsure if it's correct.
 A: If the derivatives are continuous, then the function is differentiable.
So find the derivative with respect to $x$ in $(0,0)$:
$$
\lim_{(x,y)\to(0,0)}\frac{x^2\log(x^4+y^2)-0}{x-0}
$$
and compare it with the limit of the derivative function
$$
\lim_{(x,y)\to(0,0)}f'_x(x,y)=\lim_{(x,y)\to(0,0)}\left(2x\log(x^4+y^2)+\frac{4x^5}{x^4+y^2}\right)
$$
Do the same for $f'_y(x,y)$.
A: It is sufficient to show that $f$ is continuous in a neighbourhoud of $\vec 0$ and that $f$ has continuous partial derivatives in a neighbourhoud of $\vec 0$.
Obviously, the only point where a problem could arise is in $\vec 0$. Since you showed that $f$ is continuous in this point, you only need to check that $$\frac{\delta f}{\delta x}$$ and $$\frac{\delta f}{\delta y}$$
are continuous in a neighbourhood of $\vec 0$, which isn't hard.
A: $t\log t^4=4t\log t\to 0$ as $t\to 0_+$  is a standard limit from high school.
A: This is indeed the way to go, to compute that limit, try using L'Hopital's rule on: $$ \frac{log(t^n)}{(\frac1t)} $$ with $t$ tending to zero from above.
Hopefully this should suffice for you to complete the question. 
