Let us consider a variable $x$ that changes its value from an initial state to a new state.

The difference between the new value and the original value is defined as the increment of $x$. We write this as $\Delta x$. It represents a small, finite change in the variable and can be positive, negative, or zero.

We know that the Power Rule is used when we differentiate a variable raised to a constant exponent.

If our function is $y = x^n$, where $n$ is any real number, then the derivative of $y$ with respect to $x$ is given by:
$$\frac{dy}{dx} = n x^{n-1}$$

Let us write down the expression and apply the Sum and Difference Rules. This allows us to differentiate each term separately:
$$\frac{d}{dx}(5x^2 - 3x) = \frac{d}{dx}(5x^2) - \frac{d}{dx}(3x)$$
We know that we can bring the constant coefficients to the front. Therefore, we write:
$$= 5 \cdot \frac{d}{dx}(x^2) - 3 \cdot \frac{d}{dx}(x^1)$$
Applying the Power Rule to each term gives us:
$$= 5(2x) - 3(1) = 10x - 3$$

Since $x$ and $y$ are given as separate functions of a parameter $t$, let us use Parametric Differentiation.

First, let us differentiate both equations with respect to $t$:
$$\frac{dx}{dt} = \frac{d}{dt}(t^2) = 2t$$
$$\frac{dy}{dt} = \frac{d}{dt}(2t) = 2$$
We know that the formula for parametric derivatives is:
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$
Therefore, substituting our calculated values into this ratio gives:
$$\frac{dy}{dx} = \frac{2}{2t} = \frac{1}{t}$$

First, let us rewrite the fraction with a negative exponent to make it easier to work with:
$$y = x^{-1}$$
Now, let us apply the Power Rule, using $n = -1$. This gives us:
$$\frac{dy}{dx} = -1 \cdot x^{-1 - 1} = -1 \cdot x^{-2}$$
Therefore, rewriting this with a positive exponent in the denominator yields our final result:
$$\frac{dy}{dx} = -\frac{1}{x^2}$$

Let us first find the general derivative of the function $f(x) = x^2$.

Using the Power Rule, we find:
$$f'(x) = 2x$$
Now, we wish to find the value of this derivative at the specific point where $x = 3$. Therefore, let us substitute $x = 3$ into our derivative equation:
$$f'(3) = 2(3) = 6$$

We know that the slope of a curve at any point is represented by its derivative, $\frac{dy}{dx}$.

First, let us differentiate our function:
$$\frac{dy}{dx} = 2x$$
To find the slope at the origin $(0,0)$, let us substitute the x-coordinate ($x = 0$) into our derivative:
$$\text{Slope} = 2(0) = 0$$

No, not every continuous function is derivable. While a function must be continuous to be derivable, continuity on its own is not enough.

For example, the absolute value function $f(x) = |x|$ is continuous everywhere, including at $x = 0$. However, it is not derivable at $x = 0$ because its graph has a sharp corner there, causing the left-hand and right-hand limits of the derivative to be different ($-1$ and $1$ respectively).

Let us use the Chain Rule. We can treat $u = x+a$ as our inner function, so our equation is $y = u^n$.

First, we differentiate the outer power with respect to $u$, which gives:
$$\frac{dy}{du} = n u^{n-1} = n(x+a)^{n-1}$$
Next, let us differentiate the inner function with respect to $x$:
$$\frac{du}{dx} = \frac{d}{dx}(x+a) = 1 + 0 = 1$$
Therefore, multiplying these derivatives together yields:
$$\frac{dy}{dx} = n(x+a)^{n-1} \cdot 1 = n(x+a)^{n-1}$$

First, let us rewrite the square root using a fractional exponent:
$$y = (4-x)^{1/2}$$
Now, let us apply the Chain Rule. We differentiate the outer power and then multiply by the derivative of the inside function:
$$y' = \frac{1}{2}(4-x)^{\frac{1}{2} - 1} \cdot \frac{d}{dx}(4-x)$$
We know that the derivative of the inside term is $\frac{d}{dx}(4-x) = 0 - 1 = -1$. Substituting this in gives:
$$y' = \frac{1}{2}(4-x)^{-1/2} \cdot (-1)$$
Therefore, simplifying the expression and writing it back in root form yields:
$$y' = -\frac{1}{2\sqrt{4-x}}$$

Let us write down our function as $f(x) = x^3$. To find its derivative from first principles, we must use the limit definition.

Step 1 — Find the incremented value:
Let us introduce a small increment $h$ to $x$. This gives us:
$$f(x+h) = (x+h)^3$$

Step 2 — Set up the limit definition:
We know that the first principles formula is:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
Therefore, substituting our expressions into the formula yields:
$$f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h}$$

Step 3 — Expand the binomial expression:
Using the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, we expand the term:
$$f'(x) = \lim_{h \to 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3) - x^3}{h}$$

Step 4 — Simplify and factor out h:
Let us subtract the $x^3$ terms to simplify the numerator:
$$f'(x) = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}$$
Now, let us factor out $h$ from each term in the numerator:
$$f'(x) = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h}$$
Since $h \neq 0$, we can cancel the $h$ terms:
$$f'(x) = \lim_{h \to 0} (3x^2 + 3xh + h^2)$$

Step 5 — Apply the limit:
Finally, let us substitute $h = 0$ into the remaining terms:
$$f'(x) = 3x^2 + 3x(0) + 0^2 = 3x^2$$

Let us write our function as $f(x) = \sqrt{x}$. To find the derivative using first principles, we will need to resolve a square root limit by rationalizing.

Step 1 — Find the incremented value:
Let us introduce a small increment $h$ to the variable $x$. This gives:
$$f(x+h) = \sqrt{x+h}$$

Step 2 — Set up the limit:
We know that the derivative formula is:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$$

Step 3 — Rationalize the numerator:
While solving, let us multiply both the numerator and the denominator by the conjugate expression $(\sqrt{x+h} + \sqrt{x})$:
$$f'(x) = \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}$$
Using the identity $(a-b)(a+b) = a^2 - b^2$, we can simplify the numerator:
$$f'(x) = \lim_{h \to 0} rac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}$$
$$f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$$

Step 4 — Cancel h and evaluate the limit:
Let us cancel the common factor $h$ from both the top and the bottom:
$$f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$$
Now, let us apply the limit by substituting $h = 0$:
$$f'(x) = \frac{1}{\sqrt{x+0} + \sqrt{x}} = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$$

Let us differentiate the given equation term-by-term with respect to $x$.

First, let us write down the derivative of the entire equation:
$$\frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(10)$$
Let us solve each term individually:
1. For $x^2$, the derivative is straightforward: $2x$.
2. For $xy$, we must use the Product Rule. This gives us: $x \frac{dy}{dx} + y \cdot 1$.
3. For $y^2$, we must use the Chain Rule, which yields: $2y \frac{dy}{dx}$.
4. For the constant $10$, the derivative is $0$.

Therefore, by substituting these back, we get:
$$2x + x \frac{dy}{dx} + y + 2y \frac{dy}{dx} = 0$$
Now, let us group all terms containing $\frac{dy}{dx}$ on one side, and move the remaining terms to the other side:
$$x \frac{dy}{dx} + 2y \frac{dy}{dx} = -2x - y$$
Let us factor out $\frac{dy}{dx}$ from the left-hand side:
$$(x + 2y)\frac{dy}{dx} = -(2x + y)$$
Dividing both sides by $(x + 2y)$ gives our final result:
$$\frac{dy}{dx} = -\frac{2x+y}{x+2y}$$

Let us break this composite function down into simpler parts. We can define our inner function as $u = 3x^2 + 2$.

Therefore, we can rewrite our original function as:
$$y = u^4$$
Now, let us differentiate both equations individually. First, we find the derivative of $y$ with respect to $u$ using the Power Rule:
$$\frac{dy}{du} = 4u^3 = 4(3x^2 + 2)^3$$
Next, let us differentiate $u$ with respect to $x$:
$$\frac{du}{dx} = \frac{d}{dx}(3x^2 + 2) = 6x + 0 = 6x$$
We know that the Chain Rule states:
$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$
Therefore, by multiplying our two derivative expressions together, we get:
$$\frac{dy}{dx} = 4(3x^2 + 2)^3 \cdot 6x$$
While simplifying, we multiply the constant coefficients to find the final derivative:
$$\frac{dy}{dx} = 24x(3x^2 + 2)^3$$

Since $x$ and $y$ are expressed in terms of the parameter $\theta$, let us perform Parametric Differentiation.

Step 1 — Differentiate x with respect to θ:
Using the Chain Rule on the term $\cos^3 \theta$, we get:
$$\frac{dx}{d\theta} = a \cdot 3\cos^2 \theta \cdot \frac{d}{d\theta}(\cos \theta) = 3a\cos^2 \theta(-\sin \theta) = -3a\cos^2 \theta \sin \theta$$

Step 2 — Differentiate y with respect to θ:
Using the Chain Rule on the term $\sin^3 \theta$, we get:
$$\frac{dy}{d\theta} = a \cdot 3\sin^2 \theta \cdot \frac{d}{d\theta}(\sin \theta) = 3a\sin^2 \theta(\cos \theta) = 3a\sin^2 \theta \cos \theta$$

Step 3 — Apply the parametric derivative formula:
We know that the derivative is given by the ratio of these rates:
$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$$
Therefore, substituting our expressions yields:
$$\frac{dy}{dx} = \frac{3a\sin^2 \theta \cos \theta}{-3a\cos^2 \theta \sin \theta}$$
Let us simplify by canceling the common terms $3a$, one factor of $\sin \theta$, and one factor of $\cos \theta$:
$$\frac{dy}{dx} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta$$

Let us identify our numerator as $u = x^2 + 1$ and our denominator as $v = x^2 - 1$.

First, let us find their individual derivatives with respect to $x$:
$$\frac{du}{dx} = 2x \quad \text{and} \quad \frac{dv}{dx} = 2x$$
We know that the Quotient Rule formula is:
$$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$
Therefore, substituting our expressions into the formula gives:
$$\frac{dy}{dx} = \frac{(x^2 - 1)(2x) - (x^2 + 1)(2x)}{(x^2 - 1)^2}$$
While solving, let us factor out the common term $2x$ in the numerator to simplify the algebra:
$$\frac{dy}{dx} = \frac{2x \left[ (x^2 - 1) - (x^2 + 1) \right]}{(x^2 - 1)^2}$$
Let us simplify the terms inside the square brackets:
$$(x^2 - 1 - x^2 - 1) = -2$$
Substituting this back into the numerator gives:
$$\frac{dy}{dx} = \frac{2x(-2)}{(x^2 - 1)^2} = -\frac{4x}{(x^2 - 1)^2}$$

To make this problem simple, let us introduce a new variable. Let us assume that $u = x^2$ is our independent variable.

Now, we can rewrite our original function in terms of $u$:
$$y = (1+u)^2$$
The question asks us to differentiate our function with respect to $x^2$. This means we need to find $\frac{dy}{du}$.

Let us differentiate $y$ directly with respect to $u$ using the Chain Rule:
$$\frac{dy}{du} = 2(1+u)^1 \cdot \frac{d}{du}(1+u)$$
Since the derivative of $(1+u)$ with respect to $u$ is $1$, we get:
$$\frac{dy}{du} = 2(1+u)$$
Therefore, substituting $u = x^2$ back into our equation yields:
$$\frac{dy}{du} = 2(1+x^2)$$

We know that the derivative $\frac{dy}{dx}$ gives us the slope of the tangent line at any point.

Step 1 — Find the general derivative:
$$\frac{dy}{dx} = \frac{d}{dx}(2x^2 - 3x + 1) = 4x - 3$$

Step 2 — Calculate the slope at (2, 3):
Let us substitute the x-coordinate ($x = 2$) into our derivative to find the slope $m$:
$$m = 4(2) - 3 = 8 - 3 = 5$$

Step 3 — Find the equation of the tangent line:
Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(2, 3)$ and slope $m = 5$, we write:
$$y - 3 = 5(x - 2)$$
Let us expand and simplify this equation:
$$y - 3 = 5x - 10$$
$$5x - y - 7 = 0$$

Let us define our inner function as $u = \sqrt{x} = x^{1/2}$.

Therefore, we can rewrite our function in a simpler form:
$$y = e^u$$
Now, let us find the derivative of each part:
$$\frac{dy}{du} = e^u = e^{\sqrt{x}}$$
$$\frac{du}{dx} = \frac{1}{2\sqrt{x}}$$
We know that the Chain Rule states:
$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$
Therefore, by substituting our derivatives into the formula, we get:
$$\frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}}}{2\sqrt{x}}$$

Let us write down our function as $f(x) = x^{-2} = \frac{1}{x^2}$.

Step 1 — Find the incremented value:
Let us introduce a small increment $h$ to $x$:
$$f(x+h) = \frac{1}{(x+h)^2}$$

Step 2 — Set up the limit:
We know that the first principles definition is:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$$

Step 3 — Find a common denominator:
Let us simplify the fractional terms in the numerator:
$$f'(x) = \lim_{h \to 0} \frac{1}{h} \cdot \left[ \frac{x^2 - (x+h)^2}{x^2(x+h)^2} \right]$$
Let us expand $(x+h)^2$ in the numerator:
$$f'(x) = \lim_{h \to 0} \frac{1}{h} \cdot \left[ \frac{x^2 - (x^2 + 2xh + h^2)}{x^2(x+h)^2} \right] = \lim_{h \to 0} \frac{1}{h} \cdot \left[ \frac{-2xh - h^2}{x^2(x+h)^2} \right]$$

Step 4 — Factor out h and simplify:
$$\Delta y = -h(2x+h)$$
$$f'(x) = \lim_{h \to 0} \frac{-h(2x + h)}{h \cdot x^2(x+h)^2}$$
Canceling the common factor $h$ gives:
$$f'(x) = \lim_{h \to 0} \frac{-(2x + h)}{x^2(x+h)^2}$$

Step 5 — Apply the limit:
Let us substitute $h = 0$ into our expression:
$$f'(x) = \frac{-(2x + 0)}{x^2(x+0)^2} = \frac{-2x}{x^2 \cdot x^2} = \frac{-2x}{x^4} = -\frac{2}{x^3}$$