The problem asks us to find $\frac{dy}{dx}$ for three different functions. i. $y = x^2 - 3x + 4$, using first principles. ii. $y = \frac{4x^2 - x + 3x^{-1}}{\sqrt{x}}$ iii. $y = e^{2x} \sin 2x$

1. Problem Description

The problem asks us to find $\frac{dy}{dx}$ for three different functions.

i. $y = x^2 - 3x + 4$, using first principles.

ii. $y = \frac{4x^2 - x + 3x^{-1}}{\sqrt{x}}$

iii. $y = e^{2x} \sin 2x$

2. Solution Steps

i. $y = x^2 - 3x + 4$

Using first principles, we have:

$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$f(x+h) = (x+h)^2 - 3(x+h) + 4 = x^2 + 2xh + h^2 - 3x - 3h + 4$

$f(x) = x^2 - 3x + 4$

$f(x+h) - f(x) = (x^2 + 2xh + h^2 - 3x - 3h + 4) - (x^2 - 3x + 4) = 2xh + h^2 - 3h$

$\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2 - 3h}{h} = 2x + h - 3$

$\frac{dy}{dx} = \lim_{h \to 0} (2x + h - 3) = 2x - 3$

ii. $y = \frac{4x^2 - x + 3x^{-1}}{\sqrt{x}} = \frac{4x^2 - x + \frac{3}{x}}{x^{\frac{1}{2}}} = \frac{4x^2}{x^{\frac{1}{2}}} - \frac{x}{x^{\frac{1}{2}}} + \frac{3x^{-1}}{x^{\frac{1}{2}}} = 4x^{\frac{3}{2}} - x^{\frac{1}{2}} + 3x^{-\frac{3}{2}}$

$\frac{dy}{dx} = 4 \cdot \frac{3}{2} x^{\frac{1}{2}} - \frac{1}{2} x^{-\frac{1}{2}} + 3 \cdot (-\frac{3}{2}) x^{-\frac{5}{2}} = 6x^{\frac{1}{2}} - \frac{1}{2}x^{-\frac{1}{2}} - \frac{9}{2} x^{-\frac{5}{2}} = 6\sqrt{x} - \frac{1}{2\sqrt{x}} - \frac{9}{2x^{\frac{5}{2}}}$

$\frac{dy}{dx} = 6\sqrt{x} - \frac{1}{2\sqrt{x}} - \frac{9}{2x^2\sqrt{x}} = \frac{12x^3 - x^2 - 9}{2x^2\sqrt{x}}$

iii. $y = e^{2x} \sin 2x$

Using the product rule: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$

$u = e^{2x}$, $v = \sin 2x$

$\frac{du}{dx} = 2e^{2x}$, $\frac{dv}{dx} = 2\cos 2x$

$\frac{dy}{dx} = e^{2x}(2\cos 2x) + (\sin 2x)(2e^{2x}) = 2e^{2x}\cos 2x + 2e^{2x}\sin 2x = 2e^{2x}(\cos 2x + \sin 2x)$

3. Final Answer

i. $\frac{dy}{dx} = 2x - 3$

ii. $\frac{dy}{dx} = 6\sqrt{x} - \frac{1}{2\sqrt{x}} - \frac{9}{2x^{\frac{5}{2}}} = \frac{12x^3 - x^2 - 9}{2x^2\sqrt{x}}$

iii. $\frac{dy}{dx} = 2e^{2x}(\cos 2x + \sin 2x)$