ALAVANYO SECONDARY TECHNICAL SCHOOL END OF FIRST SEMESTER, 2025 ELECTIVE MATHEMATICS: TWC 3 Hours---SECTION A [48 Marks] Answer All The Questions In This Section. 1. Find From The First Principles The Derivative Of Y = 3 X 3 + X Y = 3x^3 + X Y = 3 X 3 + X .2. If

ALAVANYO SECONDARY TECHNICAL SCHOOL END OF FIRST SEMESTER, 2025 ELECTIVE MATHEMATICS: TWC 3 Hours---SECTION A

48 marks

Answer all the questions in this section.

1. Find from the first principles the derivative of $y = 3x^3 + x$

To find the derivative of the given function $y = 3x^3 + x$ from the first principles, we will use the definition of a derivative. The derivative of a function $y = f(x)$ is defined as:

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

In this case, we have $f(x) = 3x^3 + x$. We will substitute this into the definition of a derivative and simplify.

First, let's find $f(x + h)$ by substituting $x + h$ into the function:

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

Expanding the cubic term, we get:

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

Simplifying further, we get:

$$f(x + h) = 3x^3 + 9x^2h + 9xh^2 + 3h^3 + x + h$$

Now, let's find $f(x + h) - f(x)$ by subtracting $f(x)$ from $f(x + h)$:

$$f(x + h) - f(x) = (3x^3 + 9x^2h + 9xh^2 + 3h^3 + x + h) - (3x^3 + x)$$

Simplifying further, we get:

$$f(x + h) - f(x) = 9x^2h + 9xh^2 + 3h^3 + h$$

Now, let's substitute $f(x + h) - f(x)$ and $h$ into the definition of a derivative:

$$f'(x) = \lim_{h \to 0} \frac{9x^2h + 9xh^2 + 3h^3 + h}{h}$$

Simplifying further, we get:

$$f'(x) = \lim_{h \to 0} (9x^2 + 9xh + 3h^2 + 1)$$

As $h$ approaches 0, the terms involving $h$ approach 0, and we are left with:

$$f'(x) = 9x^2 + 1$$

Therefore, the derivative of $y = 3x^3 + x$ from the first principles is $f'(x) = 9x^2 + 1$.

2. If $y = \frac{1}{x^2 + 1}$, find $\frac{dy}{dx}$ using the quotient rule.

To find $\frac{dy}{dx}$ using the quotient rule, we need to identify the numerator and denominator of the function. In this case, the numerator is 1 and the denominator is $x^2 + 1$.

The quotient rule states that if $y = \frac{f(x)}{g(x)}$, then $\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$.

In this case, we have $f(x) = 1$ and $g(x) = x^2 + 1$. We need to find $f'(x)$ and $g'(x)$.

Since $f(x) = 1$, we have $f'(x) = 0$.

To find $g'(x)$, we will use the power rule, which states that if $y = x^n$, then $y' = nx^{n-1}$.

In this case, we have $g(x) = x^2 + 1$, so $g'(x) = 2x$.

Now, let's substitute $f'(x)$, $g(x)$, $f(x)$, and $g'(x)$ into the quotient rule formula:

$$\frac{dy}{dx} = \frac{(0)(x^2 + 1) - (1)(2x)}{(x^2 + 1)^2}$$

Simplifying further, we get:

$$\frac{dy}{dx} = \frac{-2x}{(x^2 + 1)^2}$$

Therefore, the derivative of $y = \frac{1}{x^2 + 1}$ using the quotient rule is $\frac{dy}{dx} = \frac{-2x}{(x^2 + 1)^2}$.

3. Find the derivative of $y = \frac{x^2 + 1}{x^2 - 1}$ using the quotient rule.

To find the derivative of $y = \frac{x^2 + 1}{x^2 - 1}$ using the quotient rule, we need to identify the numerator and denominator of the function. In this case, the numerator is $x^2 + 1$ and the denominator is $x^2 - 1$.

The quotient rule states that if $y = \frac{f(x)}{g(x)}$, then $\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$.

In this case, we have $f(x) = x^2 + 1$ and $g(x) = x^2 - 1$. We need to find $f'(x)$ and $g'(x)$.

To find $f'(x)$, we will use the power rule, which states that if $y = x^n$, then $y' = nx^{n-1}$.

In this case, we have $f(x) = x^2 + 1$, so $f'(x) = 2x$.

To find $g'(x)$, we will also use the power rule.

In this case, we have $g(x) = x^2 - 1$, so $g'(x) = 2x$.

Now, let's substitute $f'(x)$, $g(x)$, $f(x)$, and $g'(x)$ into the quotient rule formula:

$$\frac{dy}{dx} = \frac{(2x)(x^2 - 1) - (x^2 + 1)(2x)}{(x^2 - 1)^2}$$

Simplifying further, we get:

$$\frac{dy}{dx} = \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2 - 1)^2}$$

Simplifying further, we get:

$$\frac{dy}{dx} = \frac{-4x}{(x^2 - 1)^2}$$

Therefore, the derivative of $y = \frac{x^2 + 1}{x^2 - 1}$ using the quotient rule is $\frac{dy}{dx} = \frac{-4x}{(x^2 - 1)^2}$.

4. Find the derivative of $y = \frac{1}{x^2 - 4}$ using the quotient rule.

To find the derivative of $y = \frac{1}{x^2 - 4}$ using the quotient rule, we need to identify the numerator and denominator of the function. In this case, the numerator is 1 and the denominator is $x^2 - 4$.

The quotient rule states that if $y = \frac{f(x)}{g(x)}$, then $\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$.

In this case, we have $f(x) = 1$ and $g(x) = x^2 - 4$. We need to find $f'(x)$ and $g'(x)$.

Since $f(x) = 1$, we have $f'(x) = 0$.

To find $g'(x)$, we will use the power rule, which states that if $y = x^n$, then $y' = nx^{n-1}$.

In this case, we have $g(x) = x^2 - 4$, so $g'(x) = 2x$.

Now, let's substitute $f'(x)$, $g(x)$, $f(x)$, and $g'(x)$ into the quotient rule formula:

$$\frac{dy}{dx} = \frac{(0)(x^2 - 4) - (1)(2x)}{(x^2 - 4)^2}$$

Simplifying further, we get:

$$\frac{dy}{dx} = \frac{-2x}{(x^2 - 4)^2}$$

Therefore, the derivative of $y = \frac{1}{x^2 - 4}$ using the quotient rule is $\frac{dy}{dx} = \frac{-2x}{(x^2 - 4)^2}$.

5. Find the derivative of $y = \frac{x^2 - 4}{x

ALAVANYO SECONDARY TECHNICAL SCHOOL END OF FIRST SEMESTER, 2025 ELECTIVE MATHEMATICS: TWC 3 Hours---SECTION A

48 marks

Answer all the questions in this section.

6. Find the derivative of $y = \frac{x^2 - 4}{x}$ using the quotient rule.

To find the derivative of $y = \frac{x^2 - 4}{x}$ using the quotient rule, we need to identify the numerator and denominator of the function. In this case, the numerator is $x^2 - 4$ and the denominator is $x$.

The quotient rule states that if $y = \frac{f(x)}{g(x)}$, then $\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$.

In this case, we have $f(x) = x^2 - 4$ and $g(x) = x$. We need to find $f'(x)$ and $g'(x)$.

To find $f'(x)$, we will use the power rule, which states that if $y = x^n$, then $y' = nx^{n-1}$.

In this case, we have $f(x) = x^2 - 4$, so $f'(x) = 2x$.

To find $g'(x)$, we will also use the power rule.

In this case, we have $g(x) = x$, so $g'(x) = 1$.

Now, let's substitute $f'(x)$, $g(x)$, $f(x)$, and $g'(x)$ into the quotient rule formula:

$$\frac{dy}{dx} = \frac{(2x)(x) - (x^2 - 4)(1)}{x^2}$$

Simplifying further, we get:

$$\frac{dy}{dx} = \frac{2x^2 - x^2 + 4}{x^2}$$

Simplifying further, we get:

$$\frac{dy}{dx} = \frac{x^2 + 4}{x^2}$$

Therefore, the derivative of $y = \frac{x^2 - 4}{x}$ using the quotient rule is $\frac{dy}{dx} = \frac{x^2 + 4}{x^2}$.

7. Find the derivative of $y = \frac{1}{x^2 - 4}$ using the chain rule.

To find the derivative of $y = \frac{1}{x^2 - 4}$ using the chain rule, we need to identify the outer and inner functions.

In this case, the outer function is $f(u) = \frac{1}{u}$ and the inner function is $u = x^2 - 4$.

The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

In this case, we have $f(u) = \frac{1}{u}$ and $g(x) = x^2 - 4$. We need to find $f'(u)$ and $g'(x)$.

To find $f'(u)$, we will use the power rule, which states that if $y = x^n$, then $y' = nx^{n-1}$.

In this case, we have $f(u) = \frac{1}{u}$, so $f'(u) = -\frac{1}{u^2}$.

To find $g'(x)$, we will use the power rule.

In this case, we have $g(x) = x^2 - 4$, so $g'(x) = 2x$.

Now, let's substitute $f'(u)$ and $g'(x)$ into the chain rule formula:

$$\frac{dy}{dx} = f'(g(x)) \cdot g'(x) = -\frac{1}{(x^2 - 4)^2} \cdot 2x$$

Simplifying further, we get:

$$\frac{dy}{dx} = -\frac{2x}{(x^2 - 4)^2}$$

Therefore, the derivative of $y = \frac{1}{x^2 - 4}$ using the chain rule is $\frac{dy}{dx} = -\frac{2x}{(x^2 - 4)^2}$.

8. Find the derivative of $y = \frac{x^2 - 4}{x^2 + 4}$ using the quotient rule.

To find the derivative of $y = \frac{x^2 - 4}{x^2 + 4}$ using the quotient rule, we need to identify the numerator and denominator of the function. In this case, the numerator is $x^2 - 4$ and the denominator is $x^2 + 4$.

The quotient rule states that if $y = \frac{f(x)}{g(x)}$, then $\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$.

In this case, we have $f(x) = x^2 - 4$ and $g(x) = x^2 + 4$. We need to find $f'(x)$ and $g'(x)$.

To find $f'(x)$, we will use the power rule, which states that if $y = x^n$, then $y' = nx^{n-1}$.

In this case, we have $f(x) = x^2 - 4$, so $f'(x) = 2x$.

To find $g'(x)$, we will also use the power rule.

In this case, we have $g(x) = x^2 + 4$, so $g'(x) = 2x$.

Now, let's substitute $f'(x)$, $g(x)$, $f(x)$, and $g'(x)$ into the quotient rule formula:

$$\frac{dy}{dx} = \frac{(2x)(x^2 + 4) - (x^2 - 4)(2x)}{(x^2 + 4)^2}$$

Simplifying further, we get:

$$\frac{dy}{dx} = \frac{2x^3 + 8x - 2x^3 + 8x}{(x^2 + 4)^2}$$

Simplifying further, we get:

$$\frac{dy}{dx} = \frac{16x}{(x^2 + 4)^2}$$

Therefore, the derivative of $y = \frac{x^2 - 4}{x^2 + 4}$ using the quotient rule is $\frac{dy}{dx} = \frac{16x}{(x^2 + 4)^2}$.

9. Find the derivative of $y = \frac{1}{x^2 + 4}$ using the chain rule.

To find the derivative of $y = \frac{1}{x^2 + 4}$ using the chain rule, we need to identify the outer and inner functions.

In this case, the outer function is $f(u) = \frac{1}{u}$ and the inner function is $u = x^2 + 4$.

The chain rule states that if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

In this case, we have $f(u) = \frac{1}{u}$ and $g(x) = x^2 + 4$. We need to find $f'(u)$ and $g'(x)$.

To find $f'(u)$, we will use the power rule, which states that if $y = x^n$, then $y' = nx^{n-1}$.

In this case, we have $f(u) = \frac{1}{u}$, so $f'(u) = -\frac{1}{u^2}$.

To find $g'(x)$, we will use the power rule.

In this case, we have $g(x) = x^2 + 4$, so $g'(x) = 2x$.

Now, let's substitute $f'(u)$ and $g'(x)$ into the chain rule formula:

$$\frac{dy}{dx} = f'(g(x)) \cdot g'(x) = -\frac{1}{(x^2 + 4)^2} \cdot 2x$$

Simplifying further, we get:

$$\frac{dy}{dx} = -\frac{2x}{(x^2 + 4)^2}$$

Therefore, the derivative of $y = \frac{1}{x^2 + 4}$ using the chain rule is $\frac{dy}{dx} = -\frac{2x}{(x^2 + 4)^2}$.

10. Find the derivative of $y = \frac{x^2 + 4}{x^2 - 4}$ using the quotient rule.

To find the derivative of $y = \frac{x^2 + 4}{x^2 - 4}$ using the quotient rule, we need to identify the numerator and denominator of the function. In this case, the numerator is $x^2