T1202s Maths N3 2.1. Differentiate the following function from the first principle: $$ f(x)=-2 t \pi-9 t^{2} x^{2}+4 t^{3} x $$ 2.2. Determine $$ \frac{d y}{d x} $$ of the following function by making use of the rules of differentiation. Att flnat answers must be whth a poshtive exponent and in a surd form whore appticatole. $$ y=\frac{2}{3 x^{4}}-\frac{1}{2 x^{3}}-4 \sqrt[3]{x}+8 $$ 2.3. Make use of differentiation to determine the co-ordinates of the turming points of the given function below. $$ f(x)=x\left(x^{2}-6 x+9\right) $$ Question 3 3.1. In the following trigonometric equation without the use of a calculator solve for $$ x $$; where $$ 0^{\circ} \leq x \leq 360^{\circ} $$ $$ \sqrt{2} \sin x=1 $$ 3.2. Prove that: $$ (\tan \alpha+\cot \alpha)^{2}=\operatorname{cosec}^{2} \alpha \cdot \sec ^{2} \alpha $$ (4)

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### 2.1 Differentiate the function from the first principle

The first principle of differentiation states that the derivative of a function $$ f(x) $$ at a point $$ x $$ is given by:

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

Given the function:

$$
f(x) = -2t\pi - 9t^2x^2 + 4t^3x
$$

We will calculate $$ f(x+h) $$:

$$
f(x+h) = -2t\pi - 9t^2(x+h)^2 + 4t^3(x+h)
$$

Now, we expand $$ f(x+h) $$:

$$
f(x+h) = -2t\pi - 9t^2(x^2 + 2xh + h^2) + 4t^3(x + h)
$$
$$
= -2t\pi - 9t^2x^2 - 18t^2xh - 9t^2h^2 + 4t^3x + 4t^3h
$$

Now, we can find $$ f(x+h) - f(x) $$:

$$
f(x+h) - f(x) = (-2t\pi - 9t^2x^2 - 18t^2xh - 9t^2h^2 + 4t^3x + 4t^3h) - (-2t\pi - 9t^2x^2 + 4t^3x)
$$
$$
= -18t^2xh - 9t^2h^2 + 4t^3h
$$

Now, we divide by $$ h $$:

$$
\frac{f(x+h) - f(x)}{h} = -18t^2x - 9t^2h + 4t^3
$$

Taking the limit as $$ h \to 0 $$:

$$
f'(x) = \lim_{h \to 0} (-18t^2x - 9t^2h + 4t^3) = -18t^2x + 4t^3
$$

Thus, the derivative is:

$$
f'(x) = -18t^2x + 4t^3
$$

### 2.2 Differentiate using the rules of differentiation

Given the function:

$$
y = \frac{2}{3x^4} - \frac{1}{2x^3} - 4\sqrt[3]{x} + 8
$$

We can rewrite it as:

$$
y = \frac{2}{3}x^{-4} - \frac{1}{2}x^{-3} - 4x^{1/3} + 8
$$

Now, we differentiate term by term:

1. $$ \frac{d}{dx}\left(\frac{2}{3}x^{-4}\right) = -\frac{8}{3}x^{-5} $$
2. $$ \frac{d}{dx}\left(-\frac{1}{2}x^{-3}\right) = \frac{3}{2}x^{-4} $$
3. $$ \frac{d}{dx}\left(-4x^{1/3}\right) = -\frac{4}{3}x^{-2/3} $$
4. $$ \frac{d}{dx}(8) = 0 $$

Combining these results:

$$
\frac{dy}{dx} = -\frac{8}{3}x^{-5} + \frac{3}{2}x^{-4} - \frac{4}{3}x^{-2/3}
$$

### 2.3 Find the coordinates of the turning points

Given the function:

$$
f(x) = x(x^2 - 6x + 9)
$$

First, we simplify:

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

Now, we differentiate:

$$
f'(x) = (x-3)^2 + x \cdot 2(x-3)
$$
$$
= (x-3)^2 + 2x(x-3)
$$
$$
= (x-3)((x-3) + 2x) = (x-3)(3x-3) = 3(x-3)x
$$

Setting $$ f'(x) = 0 $$:

$$
3(x-3)x = 0
$$

This gives us:

$$
x = 0 \quad \text{or} \quad x = 3
$$

Now, we find the coordinates:

1. For $$ x = 0 $$:
   $$
   f(0) = 0(0-3)^2 = 0
   $$
   So, the point is $$ (0, 0) $$.

2. For $$ x = 3 $$:
   $$
   f(3) = 3(3-3)^2 = 0
   $$
   So, the point is $$ (3, 0) $$.

### Question 3

#### 3.1 Solve the trigonometric equation

Given:

$$
\sqrt{2} \sin x = 1
$$

We can rewrite it as:

$$
\sin x = \frac{1}{\sqrt{2}} = \sin 45^\circ
$$

Thus, the solutions are:

$$
x = 45^\circ + k \cdot 360^\circ \quad \text{or} \quad x = 180^\circ - 45^\circ + k \cdot 360^\circ
$$

For $$ 0^\circ \leq x \leq 360^\circ $$:

1. $$ x = 45^\circ $$
2. $$ x = 135^\circ $$

#### 3.2 Prove the identity

We need to prove:

$$
(\tan \alpha + \cot \alpha)^2 = \csc^2 \alpha \cdot \sec^2 \alpha
$$

Starting with the left-hand side:

$$
(\tan \alpha + \cot \alpha)^2 = \left(\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha}\right)^2
$$
$$
= \left(\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha}\right)^2
$$
$$
= \left(\frac{1}{\sin \alpha \cos \alpha}\right)^2 = \frac{1}{\sin^2 \alpha \cos^2 \alpha}
$$

Now, the right-hand side:

$$
\csc^2 \alpha \cdot \sec^2 \alpha
**2.1 Differentiation from First Principle** The derivative of the function $$ f(x) = -2t\pi - 9t^2x^2 + 4t^3x $$ is: \[ f'(x) = -18t^2x + 4t^3 $$ **2.2 Differentiation Using Rules** The derivative of $$ y = \frac{2}{3x^4} - \frac{1}{2x^3} - 4\sqrt[3]{x} + 8 $$ is: $$ \frac{dy}{dx} = -\frac{8}{3}x^{-5} + \frac{3}{2}x^{-4} - \frac{4}{3}x^{-2/3} $$ **2.3 Turning Points** The turning points of $$ f(x) = x(x^2 - 6x + 9) $$ are at $$ (0, 0) $$ and $$ (3, 0) $$. **3.1 Trigonometric Equation Solution** The solutions for $$ \sqrt{2} \sin x = 1 $$ in the interval $$ 0^\circ \leq x \leq 360^\circ $$ are: $$ x = 45^\circ \quad \text{and} \quad x = 135^\circ $$ **3.2 Trigonometric Identity Proof** $$ (\tan \alpha + \cot \alpha)^2 = \csc^2 \alpha \cdot \sec^2 \alpha $$ Proof: $$ (\tan \alpha + \cot \alpha)^2 = \left(\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha}\right)^2 = \frac{1}{\sin^2 \alpha \cos^2 \alpha} = \csc^2 \alpha \cdot \sec^2 \alpha $$

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