#### Answer

$$0.0375$$

#### Work Step by Step

Given $$y=\frac{3-\sqrt{x}}{\sqrt{x+3}}, \quad a=1, \quad d x=-0.1$$
Since
\begin{align*}
f'(x)&= \frac{\frac{d}{dx}\left(3-\sqrt{x}\right)\sqrt{x+3}-\frac{d}{dx}\left(\sqrt{x+3}\right)\left(3-\sqrt{x}\right)}{\left(\sqrt{x+3}\right)^2}\\
&= \frac{-3\sqrt{x}-3}{2\sqrt{x}\left(x+3\right)\sqrt{x+3}}\\
f'(1)&= - 0.375
\end{align*}
Then
\begin{align*}
\Delta y \approx &= f'(1)dx\\
&= (-0.375)(-0.1)\\
&=0.0375
\end{align*}