Find dy/dx if y = (4x2 − 3x) / (8sqrt(x)) and y = cbrt(3x+4)

$$y = \frac{4x^2}{8\sqrt{x}} - \frac{3x}{8\sqrt{x}}$$ $$y = \frac{4x^2}{8x^{1/2}} - \frac{3x}{8x^{1/2}}$$ $$y = \frac{1}{2}x^{2 - 1/2} - \frac{3}{8}x^{1 - 1/2}$$ $$y = \frac{1}{2}x^{3/2} - \frac{3}{8}x^{1/2}$$

$$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{2}x^{3/2}\right) - \frac{d}{dx}\left(\frac{3}{8}x^{1/2}\right)$$ $$\frac{dy}{dx} = \frac{1}{2} \left(\frac{3}{2}\right) x^{\frac{3}{2} - 1} - \frac{3}{8} \left(\frac{1}{2}\right) x^{\frac{1}{2} - 1}$$ $$\frac{dy}{dx} = \frac{3}{4}x^{1/2} - \frac{3}{16} x^{-1/2}$$

$$\frac{dy}{dx} = \frac{3}{4}\sqrt{x} - \frac{3}{16\sqrt{x}}$$

The derivative for part (a) is $\boxed{\frac{3}{4}\sqrt{x} - \frac{3}{16\sqrt{x}}}$.

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 derivative for part (b) is $\boxed{\frac{1}{\sqrt[3]{(3x+4)^2}}}$.