Derivative of a function with bounded bilinear functions Let $X, X_1, X_2, Y $ be normed Vector spaces , $B : X_1 \times X_2 $  a bounded bilinear function and $f_i : X \longrightarrow X_i$ differentiable. 

How do you show that the function $H:X \longrightarrow Y , H(x) := B(f_1(x),f_2(x)) $ is differentiable and how does the derivative look like?
It's the composition of differentiable functions, which implies it is differentiable. It's derivative in direction $v$ is

$$DH (x) v= B(f_1(x),Df_2(x)v) + B(Df_1(x)v, f_2(x))$$ 

which follows from the fact that the derivative of the bilinear map
$$ (v,w) \mapsto B(v,w) $$
is given by
$$ DB(v,w) (e,f) = B(v,f)+ B(e,w)$$

This latter statement follows from the definition of the derivative from the following equality:

$$B(v+e, w+f) = B(v,w) + B(v,f) + B(e,w) + B(e,f) \\
= B(v,w) + DB(v,w)(e,f) + o(e,f)$$
(see, e.g., bilinear map and differentiability, if in doubt)
It's the composition of differentiable functions, which implies it is differentiable. It's derivative in direction $v$ is

$$DH (x) v= B(f_1(x),Df_2(x)v) + B(Df_1(x)v, f_2(x))$$ 

which follows from the fact that the derivative of the bilinear map
$$ (v,w) \mapsto B(v,w) $$
is given by
$$ DB(v,w) (e,f) = B(v,f)+ B(e,w)$$

This latter statement follows from the definition of the derivative from the following equality:

$$B(v+e, w+f) = B(v,w) + B(v,f) + B(e,w) + B(e,f) \\
= B(v,w) + DB(v,w)(e,f) + o(e,f)$$
(see, e.g., bilinear map and differentiability, if in doubt)