Answer
$\frac{dZ}{dt}=\frac{-1}{\sqrt{5}}$
Work Step by Step
We are given:
$R=10$ and $X=20$
$\frac{dR}{dt}=3$ and $\frac{dX}{dt}=2$
$Z=\sqrt{R^2+X^2}$
diffrentiating above with respect to t:
$\frac{dZ}{dt}=\frac{1}{2\sqrt{(R^2+X^2)}}(2R\frac{dR}{dt}+2X\frac{dX}{dt})$
$\frac{dZ}{dt}=\frac{1}{\sqrt{(R^2+X^2)}}(R\frac{dR}{dt}+X\frac{dX}{dt})$
$\frac{dZ}{dt}=\frac{1}{\sqrt{(10^2+20^2)}}(10\times3+20\times-2)$
$\frac{dZ}{dt}=\frac{1}{10\sqrt{(5)}}(-10)$
$\frac{dZ}{dt}=\frac{-1}{\sqrt{5}}$
thus:
$\frac{dZ}{dt}=\frac{-1}{\sqrt{5}}$