# Finding an expression for the second derivative of inner product.

Let $U \subseteq_{op.} \mathbb{R}^n$ and $f,g:U\rightarrow \mathbb{R}^m$ two 2-times differentiable funcions. Define $F:U \rightarrow \mathbb{R}; F(x) = \left \langle f(x),g(x) \right \rangle$. Obtain the expression of the bilinear form ${F''}(x)(u,v)$

Well, I tried (As my teacher suggested) defining $F = B \circ \Psi$, $B:\mathbb{R}^m\times \mathbb{R}^m \rightarrow \mathbb{R}, B(y,z) = \langle y,z \rangle$, $\Psi: U \rightarrow \mathbb{R}^m \times \mathbb{R}^m, \Psi(x) = (f(x),g(x))$. Then my aim is to use the chain rule, but I'm having trouble finding the derivatives of $B$ and $\Psi$. Can someone help me? Thanks.

• For $B$ I Found $B' = \langle f'(t),g(t) \rangle +\langle f(t), g'(t) \rangle$ – user286485 Mar 14 '16 at 13:03
• I suppose you meant to write $F = B \circ \Psi$ and $B:\mathbb{R}^m \times \mathbb{R}^m \to \mathbb{R}$? – Alex Provost Mar 14 '16 at 13:16
• You're right. Thanks! – user286485 Mar 14 '16 at 16:32

## 1 Answer

Note, that $B$ is bilinear, giving that $$B'(y,z)(u,v) = \def\<#1>{\left<#1\right>}\<y,v> + \<u,z>$$ and hence, as $B'(y,z)$ is linear, $$B''(y,z)[(u_1,v_1),(u_2, v_2)] = \<u_1, v_2> + \<u_2, v_1>$$ To compute the second derivative of $\Psi$ note, that derivatives can be taken componentwise, that is, we have $$\Psi''(x)(h,k) = \bigl(f''(x)(h,k), g''(x)(h,k)\bigr)$$ Now, by the chain rule \begin{align*} F''(x)(u,v) &= (B \circ \Psi)''(x)(u,v) \\ &= (B' \circ \Psi \cdot \Psi')'(x)(u,v)\\ &= B''\bigl(\Psi(x)\bigr)\bigl(\Psi'(x)u, \Psi'(x)v\bigr) + B'\bigl(\Psi(x)\bigr)\Psi''(x)(u,v)\\ &= \<f'(x)u, g'(x)v> + \<f'(x)v, g'(x)u> + \<f(x), g''(x)(u,v)> + \<f''(x)(u,v), g(x)> \end{align*}