Find dy/dx for y = (4x2 − 3x)/sqrt(x) and y = cbrt(3x+4).

$$y = \frac{4x^2}{\sqrt{x}} - \frac{3x}{\sqrt{x}}$$ $$y = 4x^{2 - 1/2} - 3x^{1 - 1/2}$$ $$y = 4x^{3/2} - 3x^{1/2}$$

$$\frac{dy}{dx} = \frac{d}{dx}(4x^{3/2}) - \frac{d}{dx}(3x^{1/2})$$ $$\frac{dy}{dx} = 4 \left(\frac{3}{2}\right) x^{\frac{3}{2} - 1} - 3 \left(\frac{1}{2}\right) x^{\frac{1}{2} - 1}$$ $$\frac{dy}{dx} = 6x^{1/2} - \frac{3}{2} x^{-1/2}$$

$$\frac{dy}{dx} = 6\sqrt{x} - \frac{3}{2\sqrt{x}}$$

The net force on $q_1$ is $\boxed{6\sqrt{x} - \frac{3}{2\sqrt{x}}}$.

Here, $u = 3x+4$ and $n = \frac{1}{3}$.

$$\frac{dy}{dx} = \frac{1}{3} (3x+4)^{\frac{1}{3} - 1} \cdot \frac{d}{dx}(3x+4)$$ $$\frac{dy}{dx} = \frac{1}{3} (3x+4)^{-2/3} \cdot 3$$

$$\frac{dy}{dx} = \frac{1}{(3x+4)^{2/3}}$$ $$\frac{dy}{dx} = \frac{1}{\sqrt[3]{(3x+4)^2}}$$

The net force on $q_2$ is $\boxed{\frac{1}{\sqrt[3]{(3x+4)^2}}}$.