#### Answer

$$f_x = \frac{3}{2\sqrt{3x + 4t}} .$$
$$f_t = \frac{2}{\sqrt{3x + 4t}} .$$

#### Work Step by Step

To find $f_x$, we treat $t$ as a constant and differentiate with respect to $x$. Thus
$$f_x = \frac{1}{2}(3x + 4t)^{-1/2} \cdot \frac{\delta}{\delta x}(3x + 4t) = \frac{3}{\sqrt{3x + 4t}} .$$
Similarly, to find $f_t$, we treat $x$ as constant and differentiate with respect to $t$. Thus
$$f_t = \frac{1}{2}(3x + 4t)^{-1/2} \frac{\delta}{\delta t}(3x + 4t) = \frac{4}{2\sqrt{3x + 4t}} = \frac{2}{\sqrt{3x + 4t}}.$$