Namo amitabha.
Excuse me ... .

$\displaystyle\tan\theta=\frac{h}{x}$ ... .
$\displaystyle\sec^2\theta\frac{d\theta}{dt}=-\frac{h}{x^2}\frac{dx}{dt}$ ... .
$\displaystyle\frac{d\theta}{dt}=-\frac{h}{x^2}\frac{dx}{dt}\cos^2\theta=-\frac{h}{x^2}\frac{dx}{dt}\frac{x^2}{x^2+h^2}=-\frac{h}{x^2+h^2}\frac{dx}{dt}$ ... .
http://www.thescienceforum.com/mathematics/33957-related-rates.html#post399151Allahu Akbar.