Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}, (x,y) \rightarrow 3x - x^3 - 2y^2 - y^4$. Find the first order Taylor expansion $T^1_a$ in $a = (\frac{1}{2}, \frac{1}{2})$
The main problem is that teacher want us to use the total derivative and not any partial derivatives (it's a pure math class).
If we use partial derivatives and Jocabian matrix we easily find :
$f(a) = \frac{15}{16}$
$\frac{\partial{f(a)}}{\partial{x}} = \frac{9}{4}$
$\frac{\partial{f(a)}}{\partial{y}} = \frac{-3}{2}$
So $Jf(a) = \begin{pmatrix}\frac{9}{4} & \frac{-3}{2}\end{pmatrix}$
Now, $T^1_af(x,y) = f(a) + Df(a)[x-a] = \frac{15}{16} + \begin{pmatrix}\frac{9}{4} & \frac{-3}{2}\end{pmatrix}\begin{pmatrix}x - \frac{1}{2} \\ y - \frac{1}{2} \end{pmatrix}$
We finally find $T^1_a = \frac{15}{16} + \frac{9x}{4} - \frac{3y}{2} - \frac{3}{8}$
However with total derivative :
$f(a) = \frac{15}{16}$
Let $h = (h_1, h_2) \in \mathbb{R}^2$,
$Df(a)[h] = (3 - 3x^2)h_1 - (4y + 4y^3)h_2 = (3 - \frac{3}{4})h_1 - (\frac{4}{2} + \frac{4}{8})h_2 = \frac{9}{4}x - \frac{5}{2}y$
We finally find $T^1_a = \frac{15}{16} + \frac{9x}{4} - \frac{5y}{2}$ which is not the same answer.
I'm pretty sure i'm wrong with the way I compute the total derivative, especially wrong with the way I use the vector $h$.
Could someone explain me the way it works ?
Thanks.
Edit :
I Finally found how it works with total derivative.
$f(a) = \frac{15}{16}$
Now we have $h$ which is basically the 2 directions in which we differentiate.
We could say $h_1 = x$ and $h_2 = y$.
All we have to do is differentiate $f(x,y)$ with respect to $x$ and $y$, multiply each differentiations with $h_1$ or $h_2$ and take the sum of the two differentiations.
Let $h = (h_1, h_2) \in \mathbb{R}^2$,
$Df(x)[h] = (3 - 3x^2)h_1 + (-4y - 4y^3)h_2$ $Df(a)[(x,y)] = (3 - \frac{3}{4})x + (-\frac{4}{2} - \frac{4}{8})y = \frac{9}{4}x - \frac{5}{2}y$
We finally find $T^1_a = \frac{15}{16} + \frac{9x}{4} - \frac{5y}{2}$ which is not the same answer.
Since $f(x,y)=3x-x^3-2y^2-y^4$, and since $x=\left(x-\frac12\right)+\frac12$ and $y=\left(y-\frac12\right)+\frac12$, you have\begin{multline}f(x,y)=-\left(y-\frac12\right)^4-2\left(y-\frac12\right)^3-\frac72\left(y-\frac12\right)^2-\frac52\left(y-\frac12\right)+\\-\left(x-\frac12\right)^3-\frac32\left(x-\frac12\right)^2+\frac94\left(x-\frac12\right)+\frac{13}{16}\end{multline}and therefore in order to get the Taylor expansion that you're after all you have to do is to consider the monomials whose degree is at most $2$, thereby getting:$$-\frac72\left(y-\frac12\right)^2-\frac52\left(y-\frac12\right)-\frac32\left(x-\frac12\right)^2+\frac94\left(x-\frac12\right)+\frac{13}{16}.$$