# How obtain the partial derivative of quadratic function $f(z)=\frac{1}{2} x^T(z)Q(z)x(z)$?

by Luis Mora   Last Updated August 14, 2019 09:20 AM - source

Consider $$z\in \mathbb{R}$$, $$x(z):\mathbb{R} \to \mathbb{R}^{n\times 1}$$ and $$Q(z):\mathbb{R} \to \mathbb{R}^{n\times n}$$ such that $$Q(z)=Q^T(z)>0$$. If we define the function $$f(z):\mathbb{R} \to \mathbb{R}$$ as $$f(z)=\frac{1}{2}x^T(z)Q(z)x(z)$$, thus, How obtain $$\dfrac{\partial f(z)}{\partial z}$$?

I know that $$\dfrac{\partial f(z)}{\partial z}$$ is one-dimensional, and if I derive in the traditional way, I obtain:

\begin{align} \dfrac{\partial f(z)}{\partial z} &= \left(\dfrac{\partial f(z)}{\partial x}\right)^T\dfrac{\partial x(z)}{\partial z}+ \left(\dfrac{\partial f(z)}{\partial Q}\right)^T\dfrac{\partial Q(z)}{\partial z}\\ &= \left(Q(z)x(z)\right)^T\dfrac{\partial x(z)}{\partial z}+ \left(\frac{1}{2}x(z)x^T(z)\right)^T\dfrac{\partial Q(z)}{\partial z} \end{align}

Of course I know that this solution is wrong, because the second term of right half site of previous equation is of dimension $$\mathbb{R}^{n\times n}$$

How could I get the $$\dfrac{\partial f(z)}{\partial z}$$ properly?

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I'm going to use Einstein's notation. We have that $$f=\frac{1}{2} x_iQ_{ij}x_j$$ So we have that $$f'=\frac{1}{2}\left({x'}_{i}Q_{ij}x_j+x_i{Q'}_{ij}x_j+x_iQ_{ij}{x'}_j\right)$$ Using the symmetry of $$Q$$, i.e. $$Q_{ij}=Q_{ji}$$, we get that $$f'=x_i Q_{ij} {x'}_j+\frac{1}{2} x_i {Q'}_{ij}x_j$$ i.e. $$f'=x^{t}Qx'+\frac{1}{2}x^{t}Q'x$$ Where $$x'$$ and $$Q'$$ are component-wise derivatives.