Let $f: \mathbb{R^2} \to \mathbb{R}$ with $f(x,y) := 6y+4x-x^3y$
I also have to determine the function definition $\tau_{(1,0)}(x,y)$ of the tangent plane to the graph of the function $f$ at the point $(1,0)$.
I would have taken
$f_x(x,y) = 4-3yx^2$
$f_y(x,y) = 6-x^3$
Then I would have calculated
$\nabla f(1,0) = (4, 5)$
Then we get
$$\tau_{1,0} (x,y) = f(1,0) + \nabla f(1,0) \cdot \begin{pmatrix} x \\ y\end{pmatrix} - \begin{pmatrix} 1 \\ 0\end{pmatrix} = 4 + (4 , 5) \begin{pmatrix} x - 1 \\ y-0 \end{pmatrix} = 4x + 5y $$
Therefore, the function definition $\tau_{1,0}(x,y) = 4x + 5y$
Can someone confirm this?