Given $f(x,y,z)=xy^2z^2+x$, show by first principles that f is differentiable at an arbitrary point (a,b,c).... 
Given f(x,y,z)=$xy^2z^2+x$,show by  first principles that f is differentiable at an arbitrary point (a,b,c). That is, guess the differential at that point and use the definition to show differentiability.

I know that so show differentiability at a point (a,b) it must so show that...
$$\lim_{h \to 0}  \frac{f(a+h_1,b+h_2)-f(a,b)-df(a)}{||h||}=0$$
So going a step further we should get...
$$\lim_{h \to 0}  \frac{f(a+h_1,b+h_2,c+h_3)-f(a,b,c)-df(a,b,c)}{||h||}=0$$
$$\lim_{h \to 0}  \frac{f((a+h_1)(b+h_2)^2(c+h_3)^2+(a+h_1))-(ab^2c^2+x)-(h_1+h_2+h_3)}{||h||} = 0$$
Which when h approaches zero i think becomes....
$$\frac{f((a+0)(b+0)^2(c+0)^2+(a+0))-(ab^2c^2+a)-(0+0+0)}{||h||} = 0$$
$$\frac{ab^2c^2+a-(ab^2c^2+a)}{||h||} = 0$$
$$\frac{0}{||h||} = 0$$
$\therefore$ it is differential at the arbitrary point (a,b,c). I'm confused by the other parts, specially the "guess" part. 
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Also i i think i should note the partial derivatives
$$\left( \frac{Df}{Dx}, \frac{Df}{Dy}, \frac{Df}{Dz} \right) = \left( (y^2z^2+1), 2xyz^2, 2xy^2z  \right)$$
Which means differential is....
$$\begin{bmatrix} (y^2z^2+1)&2xyz^2&2xy^2z \end{bmatrix} \begin{bmatrix} h_1\\h_2\\h_3 \end{bmatrix}  $$
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 A: You guessed right what the differential looks like; lets just call it
\begin{equation*}
Df_{(x,y,z)} = (y^2 z^2 + 1 , 2xyz^2 , 2xy^2z)
\end{equation*}
What you must check is the definition of differentiability, which for $n$ variables (that is for $f \colon \mathbb{R}^n \rightarrow \mathbb{R}$) is the following:
\begin{equation*}
\lim_{||h||\rightarrow 0} \frac{|| f(a+h) - f(a) - Df_a(h) ||}{||h||} = 0
\end{equation*}
where $a = (a_1,...,a_n)$, $h = (h_1,...,h_n)$ and $Df_a$ is the differential at $a$ (if it exists).
In your case, you must show that
\begin{equation*}
\lim_{||h||\rightarrow 0} \frac{|| f(a+h_1,b+h_2,c+h_3) - f(a,b,c)) - Df_a · (h_1,h_2,h_3) ||}{||h||} = 0
\end{equation*}
A: So how about this for the differentiability part i figure.
$$\lim_{h \to 0} \frac{f(a+h_1,b+h_2,c+h_3) - f(a,b,c) - Df(a,b,c)(h_1,h_2,h_3)}{||h||} = 0$$
At (0,0,0)
$$\lim_{h \to 0} \frac{f(0+h_1,0+h_2,0+h_3) - f(0,0,0) - Df(0,0,0)(h_1,h_2,h_3)}{||h||} = 0$$
$$\lim_{h \to 0} \frac{f(h_1,h_2,h_3) - f(0^20^2+1,000^2,00^20)*(h_1,h_2,h_3)}{||h||} = 0$$
$$\lim_{h \to 0} \frac{h_1h_2^2h_3^2+h1-(h_1)}{||h||} = 0$$
$$\lim_{h \to 0} \frac{h_1h_2^2h_3}{||h||} = 0$$
and i think $||h|| = \sqrt{h_1+h_2+h_3} $
So....
$$\lim_{h \to 0} \frac{h_1h_2^2h_3^2}{\sqrt{h_1+h_2+h_3} } = 0$$
i'm not sure how to proceed
