Find the derivative of the following functions from first principles: i) y = x2 ii) y = 3x + 2 iii) y = 4x2 + 2x + 1.

To differentiate from first principles, we use the definition of the derivative:

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

i) $y = x^2$

Step 1: Identify $f(x)$ and $f(x+h)$.

$$f(x) = x^2$$ $$f(x+h) = (x+h)^2 = x^2 + 2xh + h^2$$

Step 2: Substitute into the limit definition.

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

Step 3: Simplify the numerator.

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{2xh + h^2}{h}$$

Step 4: Factor out $h$ from the numerator and cancel.

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

Step 5: Evaluate the limit as $h \to 0$.

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

ii) $y = 3x + 2$

Step 1: Identify $f(x)$ and $f(x+h)$.

$$f(x) = 3x + 2$$ $$f(x+h) = 3(x+h) + 2 = 3x + 3h + 2$$

Step 2: Substitute into the limit definition.

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

Step 3: Simplify the numerator.

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

Step 4: Cancel $h$.

$$\frac{dy}{dx} = \lim_{h \to 0} 3$$

Step 5: Evaluate the limit.

$$\boxed{\frac{dy}{dx} = 3}$$

iii) $y = 4x^2 + 2x + 1$

Step 1: Identify $f(x)$ and $f(x+h)$.

$$f(x) = 4x^2 + 2x + 1$$ $$f(x+h) = 4(x+h)^2 + 2(x+h) + 1$$ $$f(x+h) = 4(x^2 + 2xh + h^2) + 2x + 2h + 1$$ $$f(x+h) = 4x^2 + 8xh + 4h^2 + 2x + 2h + 1$$

Step 2: Substitute into the limit definition.

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{(4x^2 + 8xh + 4h^2 + 2x + 2h + 1) - (4x^2 + 2x + 1)}{h}$$

Step 3: Simplify the numerator.

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{4x^2 + 8xh + 4h^2 + 2x + 2h + 1 - 4x^2 - 2x - 1}{h}$$ $$\frac{dy}{dx} = \lim_{h \to 0} \frac{8xh + 4h^2 + 2h}{h}$$

Step 4: Factor out $h$ from the numerator and cancel.

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{h(8x + 4h + 2)}{h}$$ $$\frac{dy}{dx} = \lim_{h \to 0} (8x + 4h + 2)$$

Step 5: Evaluate the limit as $h \to 0$.

$$\frac{dy}{dx} = 8x + 4(0) + 2$$ $$\boxed{\frac{dy}{dx} = 8x + 2}$$

iv) $y = \sin x$

Step 1: Identify $f(x)$ and $f(x+h)$.

$$f(x) = \sin x$$ $$f(x+h) = \sin(x+h)$$

Step 2: Substitute into the limit definition.

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

Step 3: Use the trigonometric identity $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.
Here $A = x+h$ and $B = x$.

$$\frac{A+B}{2} = \frac{x+h+x}{2} = \frac{2x+h}{2} = x + \frac{h}{2}$$ $$\frac{A-B}{2} = \frac{x+h-x}{2} = \frac{h}{2}$$

So, $\sin(x+h) - \sin x = 2 \cos\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$.

Step 4: Substitute back into the limit.

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{2 \cos\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}$$

Step 5: Rearrange the terms to use the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

$$\frac{dy}{dx} = \lim_{h \to 0} \left[ \cos\left(x + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right]$$

Step 6: Evaluate the limit. As $h \to 0$, $\frac{h}{2} \to 0$.

$$\frac{dy}{dx} = \cos(x + 0) \cdot 1$$ $$\boxed{\frac{dy}{dx} = \cos x}$$

v) $y = \cos x$

Step 1: Identify $f(x)$ and $f(x+h)$.

$$f(x) = \cos x$$ $$f(x+h) = \cos(x+h)$$

Step 2: Substitute into the limit definition.

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

Step 3: Use the trigonometric identity $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.
Here $A = x+h$ and $B = x$.

$$\frac{A+B}{2} = x + \frac{h}{2}$$ $$\frac{A-B}{2} = \frac{h}{2}$$

So, $\cos(x+h) - \cos x = -2 \sin\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$.

Step 4: Substitute back into the limit.

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

Step 5: Rearrange the terms to use the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

$$\frac{dy}{dx} = \lim_{h \to 0} \left[ -\sin\left(x + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right]$$

Step 6: Evaluate the limit. As $h \to 0$, $\frac{h}{2} \to 0$.

$$\frac{dy}{dx} = -\sin(x + 0) \cdot 1$$ $$\boxed{\frac{dy}{dx} = -\sin x}$$

vi) $y = \tan x$

Step 1: Identify $f(x)$ and $f(x+h)$.

$$f(x) = \tan x$$ $$f(x+h) = \tan(x+h)$$

Step 2: Substitute into the limit definition.

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

Step 3: Rewrite $\tan x = \frac{\sin x}{\cos x}$.

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

Step 4: Combine the fractions in the numerator.

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

Step 5: Use the trigonometric identity $\sin A \cos B - \cos A \sin B = \sin(A-B)$.
Here $A = x+h$ and $B = x$. So, $\sin((x+h)-x) = \sin h$.

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

Step 6: Separate the limit using $\lim_{h \to 0} \frac{\sin h}{h} = 1$.

$$\frac{dy}{dx} = \lim_{h \to 0} \left( \frac{\sin h}{h} \cdot \frac{1}{\cos(x+h)\cos x} \right)$$

Step 7: Evaluate the limit.

$$\frac{dy}{dx} = 1 \cdot \frac{1}{\cos(x+0)\cos x}$$ $$\frac{dy}{dx} = \frac{1}{\cos x \cos x} = \frac{1}{\cos^2 x}$$ $$\boxed{\frac{dy}{dx} = \sec^2 x}$$

vii) $y = x^3 + x^2 - x + 4$

Step 1: Identify $f(x)$ and $f(x+h)$.

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

Expand the terms:

$$(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$$ $$(x+h)^2 = x^2 + 2xh + h^2$$

So,

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

Step 2: Substitute into the limit definition.

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3 + x^2 + 2xh + h^2 - x - h + 4) - (x^3 + x^2 - x + 4)}{h}$$

Step 3: Simplify the numerator by canceling terms.

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

Step 4: Factor out $h$ from the numerator and cancel.

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2 + 2x + h - 1)}{h}$$ $$$$