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Let $B ∈ \mathcal L^2(\mathbb R^n, \mathbb R^m)$ and $f (\mathbf x) = B(\mathbf x, \mathbf x)$. Show that $D f (\mathbf x_0)(\mathbf h) = B(\mathbf x_0, \mathbf h) + B(\mathbf h, \mathbf x_0)$.

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Note that $f=B\circ (id,id)$, so $Df(x) = DB(id(x),id(x)) \circ (D(id)(x), D(id)(x)) = DB(x,x) \circ (id,id)$.

Thus, $Df(x)h = DB(x,x)(h,h)$.

Recall that $DB(x,y)(h,k) = B(x,k) + B(h,y)$.

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    what is your id here? – 2017-02-22
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    and how do you know $DB(x,y)(h,k)=B(x,k)+B(h,y)$? – 2017-02-22
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    @R.T. $id$ is the identity. For your second question, you can readily verify that. – 2017-02-22
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    Yeah, but what identity? – 2017-02-22
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    @R.T. assuming that by $\mathcal{L}^2(R^n,R^m)$ you mean the space of all bilinear functions $\Bbb R^n \times \Bbb R^n \to \Bbb R^m$, it's the identity of $\Bbb R^n$. – 2017-02-22
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    If I try to evaluate the derivative by using $f(x+h)-f(x)$, do you know how can I show $\lim_{h\to 0} \frac{\|B(h,h)\|}{\|h\|}=0$? Thanks. – 2017-02-23
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    @R.T. we have $\|B(h,h)\| \le \|B\| \|h\|^2$; in general one has $\|B(x,y)\| \le \|B\|\|x\|\|y\|$. – 2017-02-23