What is the time derivative at t=0 for a second order low-pass step response?
The under-damped (and frequency normalized) low-pass filter is this: -

$$H(s) = \dfrac{1}{s^2+2\zeta s+1}$$

It has a step response of: -

$$\dfrac{1}{s}\cdot\dfrac{1}{s^2+2\zeta s+1}$$

And, for the convenience of being able to perform an inverse Laplace transformation, it is made equal to this: -

$$\dfrac{1}{s[(s+a)^2+b]}$$

Where \$a=\zeta\$ and \$b = \sqrt{1-\zeta^2}\$

Using Laplace tables this converts to: -

$$1-\dfrac{1}{\sqrt{1-\zeta^2}}\cdot e^{-\zeta t}\cdot \sin(t\cdot\sqrt{1-\zeta^2}+\phi)$$

Where \$\phi = \arccos(\zeta)\$

Examining the formula you can see that when \$\zeta\$ approaches zero, the time derivative also approaches zero at t = 0. See the lower half of the picture: -



Picture source.

For other values of \$\zeta\$ just differentiate the formula above the picture but, what you will find is that the derivative at t = 0 is always zero because if you look at a typical 2nd order low- pass filter: -

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And you apply a step at the input, the rate of change of output voltage of the capacitor is proportional to current BUT the current doesn't change instantly because of the inductor. Hence the derivative at t = 0 is always zero.