The problem consists of three parts: (i) Determine if the function $f(x) = x^2$ is odd or even. (ii) Given the equation $(x + yi) = (3 + i)(2 - 3i)$, solve for $x$ and $y$. (iii) Differentiate $y = x^2$ using first principles.

1. Problem Description

The problem consists of three parts:

(i) Determine if the function $f(x) = x^2$ is odd or even.

(ii) Given the equation $(x + yi) = (3 + i)(2 - 3i)$, solve for $x$ and $y$.

(iii) Differentiate $y = x^2$ using first principles.

2. Solution Steps

(i) To determine if $f(x) = x^2$ is odd or even, we evaluate $f(-x)$.

$f(-x) = (-x)^2 = x^2 = f(x)$.

Since $f(-x) = f(x)$, the function is even.

(ii) We are given $(x + yi) = (3 + i)(2 - 3i)$.

First, expand the right-hand side:

$(3 + i)(2 - 3i) = 3(2) + 3(-3i) + i(2) + i(-3i) = 6 - 9i + 2i - 3i^2$.

Since $i^2 = -1$, we have:

$6 - 9i + 2i - 3(-1) = 6 - 7i + 3 = 9 - 7i$.

Therefore, $x + yi = 9 - 7i$.

Equating the real and imaginary parts, we get $x = 9$ and $y = -7$.

(iii) To differentiate $y = x^2$ using first principles, we use the definition of the derivative:

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

In this case, $f(x) = x^2$, so $f(x + h) = (x + h)^2 = x^2 + 2xh + h^2$.

Then,

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

We can cancel $h$ (since $h \ne 0$):

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

As $h$ approaches 0, we have:

$\frac{dy}{dx} = 2x + 0 = 2x$.

3. Final Answer

(i) The function $f(x) = x^2$ is even.

(ii) $x = 9$ and $y = -7$.

(iii) $\frac{dy}{dx} = 2x$.