Derivative of bilinear forms I want to solve the following problems:

  
*
  
*Let $f:\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ be a bilinear form.
  Prove that it's differential is $$ Df_{(x,y)}(a,b) = f(x,b) + f(a,y).$$
  
*Let $f:\mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^3$ be the cross product funtion, that is, $f(x,y) = x \times y$. Calulate it's derivate in the point $(x,y)$.
  

I know the definition of differentiability for funtions $f:\mathbb{R}^n \to \mathbb{R}^m$. But here I am working with other funtions, so I don't know how to start.
Thanks. 
 A: For the first part,
\begin{align}
Df_{(x,y)}(a,b) &= \frac{d}{dt}\bigg|_{t = 0} f((x,y) + t(a,b))\\
&= \frac{d}{dt}\bigg|_{t = 0} f(x + ta, y + tb)\\
&= \frac{d}{dt}\bigg|_{t = 0} (f(x,y) + tf(x,b) + tf(a,y) + t^2f(a,b))\\
&= f(x,b) + f(a,y).
\end{align}
The bilinearity condition was used third line.
For the second part, show that $f$ is a bilinear form, and use the result in part 1 to find the derivative of $f$ at $(x,y)$.
A: Read part IV of Example 6.127 in https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/mathematical-methods. This is the simplest and most direct and natural way to answer the two proposed questions. Part I-Part III answer some side issues raised in this webpage. The proof given in part IV consists of two steps. The first step is to express the bilinear form as a product of three factors: a row vector, the matrix of the bilinear form with respect to the standard bases, and a column vector. Next step: the desired results follow directly from the definition of differential.
