Given $f(x,y)=5+2x+4y+x^2+y^2+(x^2y^4)^\frac15$, show that it is differentiable at $(0,0)$. I was given the function:
$f(x,y)=5+2x+4y+x^2+y^2+(x^2y^4)^\frac15$
I need to show it is differentiable at $(0,0)$. 
I started using the method of differentials and infinitesimal functions:
$\Delta f=f(0+\Delta x,0+\Delta y)-f(0,0)$
$\Delta f=2\Delta x + 4 \Delta y + (\Delta x)^2 + (\Delta y)^2 + ((\Delta x)^2(\Delta y)^4)^\frac15$
Now I cannot see a way how I can obtain infinitesimal functions* $\alpha$ and $\beta$ such that:
$\Delta f=2\Delta x + 4 \Delta y + \alpha\Delta x + \beta \Delta y$, where $\alpha,\beta=o(\rho)$ 
(Where $\rho^2 = x^2+y^2$, and $o$ is the Landau little-$o$ notation defined as $\alpha=o(\rho)\iff\lim \alpha/\rho=0$). 
I managed to get 3 "infinitesimal functions": $\alpha=\Delta x$, $\beta=\Delta y$, and $\gamma=(\Delta x^2\Delta y^{-1})^\frac15$ but I am having trouble showing $\gamma=o(\rho)$ because frankly it doesn't seem to be.
Any hints? If this is not possible, are there any other methods that one would recommend?
*My prof defines infinitesimal functions as: $\alpha(x)$ is infinitesimal at $a$ if $\lim_{x\rightarrow a}\alpha(x)=0$

Another approach using the definition of differentiability:
To use the horrible limit
$$\lim_{(x,y)\to(0,0)}{\dfrac{f(x,y)-[f(0,0)+f'_x(0,0)(x-0)+f'_y(0,0)(y-0)]}{\sqrt{{(x-0)}^2+{(y-0)}^2}}}\tag 1$$
we need first to calculate the partial derivatives at $(0,0)$ using the fact that $f$ is $C^1(\text{Dom}(f))$. Let's begin.
Remember that $f(x,y)=5+2x+4y+x^2+y^2+\sqrt[5]{x^2y^4}$.
Finding the domain of the function
$\text{Dom}(f)=\mathbb R^2$ because is a sum of continuous.
Finding partial derivatives
$$\begin{array}{llc}
f'((0,0);(a,b))&&= \\
&\displaystyle\lim_{h\to 0}{\dfrac{f(ah,bh)-f(0,0)}{h}}&= \\
&\displaystyle\lim_{h\to 0}{\dfrac{\left(5+2ah+4bh+a^2h^2+b^2h^2+\sqrt[5]{a^2h^2b^4h^4}\right)-5}{h}}&= \\
&\displaystyle\lim_{h\to 0}{\dfrac{h\left(2a+4b+a^2h+b^2h+\frac{\sqrt[5]{a^2b^4h^6}}{h}\right)}{h}}&= \\
&\displaystyle\lim_{h\to 0}{\left(2a+4b+a^2h+b^2h+\sqrt[5]{a^2b^4h}\right)}&= \\
&2a+4b.
\end{array}$$
Thus the directional derivatives are continuous for all $(x,y)\in\mathbb R^2$, in particular for $(x,y)=(0,0)$, thus $f(x,y)\in C^1\left(\mathbb R^2\right)$ so
$$\begin{matrix}
f'((0,0);(1,0))&=&f'_x(0,0)&=&2\cdot 1+4\cdot 0&=&2, \\
f'((0,0);(0,1))&=&f'_y(0,0)&=&2\cdot 0+4\cdot 1&=&4.
\end{matrix}$$
Proving the differentiability
From $(1)$
$$\begin{array}{lcc}
\displaystyle\lim_{(x,y)\to(0,0)}{\dfrac{5+2x+4y+x^2+y^2+\sqrt[5]{x^2y^4}-[5+2(x-0)+4(y-0)]}{\sqrt{x^2+y^2}}} &=& \\
\displaystyle\lim_{(x,y)\to(0,0)}{\dfrac{x^2+y^2+\sqrt[5]{x^2y^4}}{\sqrt{x^2+y^2}}}&\underbrace =_{(x,\,y)\;=\;(\rho\cos\theta,\,\rho\sin\theta)}&\\
\displaystyle{\lim_{\rho\to 0^+}}_{\theta\in[0,2\pi)}{\dfrac{\rho^2\cos^2\theta+\rho^2\sin^2\theta+\sqrt[5]{\rho^2\cos^2\theta\cdot\rho^4\sin^4\theta}}{\sqrt{\rho^2\cos^2\theta+\rho^2\sin^2\theta}}}&=&\\
\displaystyle{\lim_{\rho\to 0^+}}_{\theta\in[0,2\pi)}{\dfrac{\rho^2+\rho\sqrt[5]{p\cos^2\theta\cdot\sin^4\theta}}{\rho}}&=&\\
\displaystyle{\lim_{\rho\to 0^+}}_{\theta\in[0,2\pi)}{\left(\rho+\sqrt[5]{p\cos^2\theta\cdot\sin^4\theta}\right)}&=&\\
0,&&
\end{array}$$
hence
$$\boxed{f(x,y)\quad\text{is differentiable at}\quad(0,0)}.$$
