QUESTION 4 Use the method of first principles to determine the derivative of $$ f(x)=\frac{6}{x} $$ Use an appropriate method off differentiation to determine the derivative of the following functions (sim- plify your answers as far as possible): (i) $$ f(x)=\sqrt{x}+3^{x}-\frac{2}{x^{3}} $$ (ii) $$ g(x)=(\sin x)(\ln x)(\cos x) $$ (iii) $$ h(x)=\left(x^{2}-2\right)^{x+1} $$

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Certainly! Let's tackle **Question 4** step by step.

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### **Part 1: Using the Method of First Principles to Determine the Derivative of $$ f(x) = \frac{6}{x} $$**

**First Principles Definition:**
The derivative of a function $$ f(x) $$ using first principles (the limit definition) is given by:

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

**Applying to $$ f(x) = \frac{6}{x} $$:**

1. **Compute $$ f(x+h) $$:**
   $$
   f(x+h) = \frac{6}{x + h}
   $$

2. **Set Up the Difference Quotient:**
   $$
   f'(x) = \lim_{h \to 0} \frac{\frac{6}{x + h} - \frac{6}{x}}{h}
   $$

3. **Simplify the Numerator:**
   $$
   \frac{6}{x + h} - \frac{6}{x} = 6 \left( \frac{1}{x + h} - \frac{1}{x} \right) = 6 \left( \frac{x - (x + h)}{x(x + h)} \right) = 6 \left( \frac{-h}{x(x + h)} \right) = -\frac{6h}{x(x + h)}
   $$

4. **Formulate the Difference Quotient:**
   $$
   f'(x) = \lim_{h \to 0} \frac{-\frac{6h}{x(x + h)}}{h} = \lim_{h \to 0} -\frac{6}{x(x + h)} 
   $$

5. **Take the Limit as $$ h \to 0 $$:**
   $$
   f'(x) = -\frac{6}{x \cdot x} = -\frac{6}{x^2}
   $$

**Final Answer:**
$$
f'(x) = -\frac{6}{x^2}
$$

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### **Part 2: Using Appropriate Differentiation Methods**

We'll now find the derivatives of the following functions using appropriate differentiation rules.

#### **(i) $$ f(x) = \sqrt{x} + 3^{x} - \frac{2}{x^{3}} $$**

**Rewrite the Function for Easier Differentiation:**
$$
f(x) = x^{1/2} + 3^{x} - 2x^{-3}
$$

**Differentiate Term by Term:**

1. **Derivative of $$ x^{1/2} $$:**
   $$
   \frac{d}{dx} \left( x^{1/2} \right) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}
   $$

2. **Derivative of $$ 3^{x} $$:**
   $$
   \frac{d}{dx} \left( 3^{x} \right) = 3^{x} \ln 3
   $$

3. **Derivative of $$ -2x^{-3} $$:**
   $$
   \frac{d}{dx} \left( -2x^{-3} \right) = -2 \cdot (-3) x^{-4} = 6x^{-4} = \frac{6}{x^4}
   $$

**Combine the Derivatives:**
$$
f'(x) = \frac{1}{2\sqrt{x}} + 3^{x} \ln 3 + \frac{6}{x^4}
$$

**Final Answer:**
$$
f'(x) = \frac{1}{2\sqrt{x}} + 3^{x} \ln 3 + \frac{6}{x^4}
$$

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#### **(ii) $$ g(x) = \sin x \cdot \ln x \cdot \cos x $$**

**Simplify the Function:**
Notice that $$ \sin x \cdot \cos x = \frac{1}{2} \sin(2x) $$, but for differentiation purposes, it's more straightforward to use the product rule for three functions.

**Product Rule for Three Functions:**
If $$ g(x) = u(x) \cdot v(x) \cdot w(x) $$, then:

$$
g'(x) = u'(x) \cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x)
$$

**Let:**
$$
u(x) = \sin x, \quad v(x) = \ln x, \quad w(x) = \cos x
$$

**Compute the Derivatives:**

1. **$$ u'(x) = \cos x $$**
2. **$$ v'(x) = \frac{1}{x} $$**
3. **$$ w'(x) = -\sin x $$**

**Apply the Product Rule:**
$$
g'(x) = (\cos x) \cdot (\ln x) \cdot (\cos x) + (\sin x) \cdot \left( \frac{1}{x} \right) \cdot (\cos x) + (\sin x) \cdot (\ln x) \cdot (-\sin x)
$$

**Simplify Each Term:**

1. **First Term:**
   $$
   \cos x \cdot \ln x \cdot \cos x = \ln x \cdot \cos^2 x
   $$

2. **Second Term:**
   $$
   \sin x \cdot \frac{1}{x} \cdot \cos x = \frac{\sin x \cos x}{x}
   $$

3. **Third Term:**
   $$
   \sin x \cdot \ln x \cdot (-\sin x) = -\ln x \cdot \sin^2 x
   $$

**Combine All Terms:**
$$
g'(x) = \ln x \cdot \cos^2 x + \frac{\sin x \cos x}{x} - \ln x \cdot \sin^2 x
$$

**Factor Out $$ \ln x $$ from the First and Third Terms:**
$$
g'(x) = \ln x ( \cos^2 x - \sin^2 x ) + \frac{\sin x \cos x}{x}
$$

**Use Trigonometric Identity:**
$$
\cos^2 x - \sin^2 x = \cos(2x)
$$

$$
g'(x) = \ln x \cdot \cos(2x) + \frac{\sin x \cos x}{x}
$$

**Final Answer:**
$$
g'(x) = \ln x \cdot \cos(2x) + \frac{\sin x \cos x}{x}
$$

*Alternatively, the second term can be written using the double-angle identity $$ \sin x \cos x = \frac{1}{2} \sin(2x) $$:*

$$
g'(x) = \ln x \cdot \cos(2x) + \frac{1}{2} \cdot \frac{\sin(2x)}{x}
$$

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#### **(iii) $$ h(x) = (x^{2} - 2)^{x + 1} $$**

**Approach:**
This function is of the form $$ y = u(x)^{v(x)} $$. To differentiate such functions, it's convenient to take the natural logarithm of both sides and then differentiate implicitly.

**Step-by-Step Process:**

1. **Take Natural Logarithm of Both Sides:**
   $$
   \ln y = \ln \left( (x^{2} - 2)^{x + 1} \right) = (x + 1) \ln (x^{2} - 2)
   $$

2. **Differentiate Both Sides with Respect to $$ x $$:**
   $$
   \frac{d}{dx} (\ln y) = \frac{d}{dx} \left( (x + 1) \ln (x^{2} - 2) \right )
   $$
   $$
   \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx} (x + 1) \cdot \ln (x^{2} - 2) + (x + 1) \cdot \frac{d}{dx} (\ln (x^{2} - 2))
   $$

3. **Compute Each Derivative:**

- **Derivative of $$ x + 1 $$:**
     $$
     \frac{d}{dx} (x + 1) = 1
     $$
     
   - **Derivative of $$ \ln (x^{2} - 2) $$:**
     $$
     \frac{d}{dx} \ln (x^{2} - 2) = \frac{2x}{x^{2} - 2}
     $$

4. **Substitute Back Into the Equation:**
   $$
   \frac{1}{y} \cdot \frac{dy}{dx} = 1 \cdot \ln (x^{2} - 2) + (x + 1) \cdot \frac{2x}{x^{2} - 2}
   $$
   $$
   \frac{dy}{dx} = y \left( \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right )
   $$

5. **Replace $$ y $$ with the Original Function $$ h(x) $$:**
   $$
   \frac{dy}{dx} = (x^{2} - 2)^{x + 1} \left( \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right )
   $$

6. **Simplify if Desired:**
   You can factor to combine the terms inside the parentheses, but it's generally acceptable to leave the answer in the above form.

**Final Answer:**
$$
h'(x) = (x^{2} - 2)^{x + 1} \left( \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right )
$$

*Alternatively, factor out $$ \frac{1}{x^{2} - 2} $$ inside the parentheses:*

$$
h'(x) = (x^{2} - 2)^{x + 1} \left( \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right ) = (x^{2} - 2)^{x + 1} \left( \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right )
$$

This form is already simplified appropriately.

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**Part 1:** $$ f'(x) = -\frac{6}{x^2} $$ **Part 2:** $$ f'(x) = \frac{1}{2\sqrt{x}} + 3^{x} \ln 3 + \frac{6}{x^4} $$ $$ g'(x) = \ln x \cdot \cos(2x) + \frac{\sin x \cos x}{x} $$ $$ h'(x) = (x^{2} - 2)^{x + 1} \left( \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right ) $$

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