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) \( h(x)=\left(x^{2}-2\right)^{x+1} \)

Certainly! Let's tackle **Question 4** step by step. --- ### **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:** \[ f'(x) = -\frac{6}{x^2} \] --- ### **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:** \[ f'(x) = \frac{1}{2\sqrt{x}} + 3^{x} \ln 3 + \frac{6}{x^4} \] --- #### **(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 \( \sin x \cos x = \frac{1}{2} \sin(2x) \):* \[ g'(x) = \ln x \cdot \cos(2x) + \frac{1}{2} \cdot \frac{\sin(2x)}{x} \] --- #### **(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. ---