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How would I take a derivative of a quadratic form with respect to a scalar, i.e. $$\frac{dx(t)^T\mathbf{Y}x(t)}{dt}$$ I have already tried splitting the quadratic form into its elements, and also applied the product rule, but it did not help. Any help is appreciated.

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    $\begingroup$ If you understand the product rule, it should not be too hard to see that it is $x'(t)^TYx(t)+x(t)^TYx'(t)$. $\endgroup$ – Captain Lama May 8 at 15:43
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The hint of CaptainLama is correct.

An alternative way is by expanding the form and observing what square terms

$$a\,x_0(t)^2\to 2a\,x_0(t)x_0'(t)$$

and double product terms

$$2b\,x_0(t)x_1(t)\to 2b\,x_0(t)'x_1(t)+2b\,x_0(t)x_1'(t)$$

become.

This should convince you that the derivative is the bilinear form

$$2x'(t)^TYx(t).$$

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Since derivative, transposition and dot multiplication are linear operations, you can apply product rule: $$ \frac d{dt}(x^\intercal Yx) = \frac{dx^\intercal}{dt}Yx+x^\intercal\frac{Y^\intercal}{dt}x + x^\intercal Y\frac{dx}{dt}. $$

Since $dY/dt=0$, we can remove the second term. And since the first term is scalar, we can replace it with its transpose:

$$ \frac{dx^\intercal}{dt}Yx + x^\intercal Y\frac{dx}{dt} = \left(\frac{dx^\intercal}{dt}Yx\right)^\intercal + x^\intercal Y\frac{dx}{dt} = x^\intercal Y^\intercal\left(\frac{dx^\intercal}{dt}\right)^\intercal + \frac{dx^\intercal}{dt}Yx = x^\intercal(Y+Y^\intercal)\frac{dx}{dt} $$

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