Answer
$f'(x)=-\frac{7}{(3+x)^{2}}$
Work Step by Step
$$f(x+h)=\frac{4-x-h}{3+x+h}$$
$$\frac{f(x+h)-f(x)}{h}=\frac{\frac{4-x-h}{3+x+h}-\frac{4-x}{3+x}}{h}=-\frac{7}{9+6x+x^{2}+3h+hx}$$
$$f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h \to 0}-\frac{7}{9+6x+x^{2}+3h+hx}=-\frac{7}{9+6x+x^{2}}=-\frac{7}{(3+x)^{2}}$$
