Find x'(t) given x(t) = 10 / (9e⁻²⁰t + 1).

The derivative of $x(t)$ will be $\frac{d}{dt}[10(9e^{-20t} + 1)^{-1}]$.
Let $u = 9e^{-20t} + 1$. Then $x(t) = 10u^{-1}$.
The chain rule states $\frac{dx}{dt} = \frac{dx}{du} \cdot \frac{du}{dt}$.
First, find $\frac{dx}{du}$:

$$\frac{dx}{du} = \frac{d}{du}(10u^{-1}) = 10(-1)u^{-2} = -10u^{-2}$$

$$u = 9e^{-20t} + 1$$ $$\frac{du}{dt} = \frac{d}{dt}(9e^{-20t} + 1)$$

Differentiate $9e^{-20t}$: Use the chain rule again. Let $w = -20t$, so $\frac{dw}{dt} = -20$.

$$\frac{d}{dt}(9e^{-20t}) = 9 \cdot e^{-20t} \cdot (-20) = -180e^{-20t}$$

The derivative of the constant 1 is 0.
So, $\frac{du}{dt} = -180e^{-20t}$

$$\dot{x}(t) = (-10u^{-2}) \cdot (-180e^{-20t})$$

Substitute $u = 9e^{-20t} + 1$ back into the expression.

$$\dot{x}(t) = -10(9e^{-20t} + 1)^{-2} \cdot (-180e^{-20t})$$

$$\dot{x}(t) = \frac{(-10)(-180e^{-20t})}{(9e^{-20t} + 1)^2}$$ $$\dot{x}(t) = \boxed{\frac{1800e^{-20t}}{(9e^{-20t} + 1)^2}}$$