UUESTION 4 Use an appropriate method off differentiation to determine the derivative of the following 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) \( h(x)=\left(x^{2}-2\right)^{x+1} \)

Sure, let's find the derivatives of each function step by step. --- ### **Problem (i)** \[ f(x) = \sqrt{x} + 3^{x} - \frac{2}{x^{3}} \] **Solution:** We'll differentiate each term separately. 1. **Derivative of \(\sqrt{x}\):** \[ \sqrt{x} = x^{1/2} \\ \frac{d}{dx} x^{1/2} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \] 2. **Derivative of \(3^{x}\):** The derivative of \(a^{x}\) with respect to \(x\) is \(a^{x} \ln a\). \[ \frac{d}{dx} 3^{x} = 3^{x} \ln 3 \] 3. **Derivative of \(-\frac{2}{x^{3}}\):** Rewrite the term as \(-2x^{-3}\) and differentiate. \[ \frac{d}{dx} (-2x^{-3}) = -2 \cdot (-3) x^{-4} = 6x^{-4} = \frac{6}{x^{4}} \] **Combine the derivatives:** \[ f'(x) = \frac{1}{2\sqrt{x}} + 3^{x} \ln 3 + \frac{6}{x^{4}} \] --- ### **Problem (ii)** \[ g(x) = (\sin x)(\ln x) (\cos x) \] **Solution:** First, it's easier to handle this by recognizing it as a product of three functions: \[ g(x) = \sin x \cdot \ln x \cdot \cos x \] To differentiate \(g(x)\), we'll use the **product rule** for multiple functions. The product rule for three functions states: \[ \frac{d}{dx} [u \cdot v \cdot w] = u' \cdot v \cdot w + u \cdot v' \cdot w + u \cdot v \cdot w' \] Let: \[ u = \sin x \\ v = \ln x \\ w = \cos x \] Compute each derivative: 1. **\(u' = \cos x\)** 2. **\(v' = \frac{1}{x}\)** 3. **\(w' = -\sin x\)** Apply the product rule: \[ g'(x) = (\cos x)(\ln x)(\cos x) + (\sin x)\left(\frac{1}{x}\right)(\cos x) + (\sin x)(\ln x)(-\sin x) \] Simplify each term: \[ g'(x) = \cos^2 x \cdot \ln x + \frac{\sin x \cos x}{x} - \sin^2 x \cdot \ln x \] We can factor \(\ln x\) from the first and third terms: \[ g'(x) = (\cos^2 x - \sin^2 x) \cdot \ln x + \frac{\sin x \cos x}{x} \] Alternatively, you can leave it in the expanded form as: \[ g'(x) = \cos^2 x \cdot \ln x + \frac{\sin x \cos x}{x} - \sin^2 x \cdot \ln x \] --- ### **Problem (iii)** \[ h(x) = \left( x^{2} - 2 \right)^{x + 1} \] **Solution:** This function has a variable in both the base and the exponent. To differentiate such functions, we'll use **logarithmic differentiation**. 1. **Take the natural logarithm of both sides:** \[ \ln h(x) = (x + 1) \cdot \ln (x^{2} - 2) \] 2. **Differentiate both sides with respect to \(x\):** Using the chain rule on the left and the product rule on the right. \[ \frac{h'(x)}{h(x)} = \frac{d}{dx} \left[ (x + 1) \cdot \ln (x^{2} - 2) \right] \] Apply the product rule: \[ \frac{h'(x)}{h(x)} = \frac{d}{dx}(x + 1) \cdot \ln (x^{2} - 2) + (x + 1) \cdot \frac{d}{dx} \ln (x^{2} - 2) \] Compute each derivative: - \(\frac{d}{dx}(x + 1) = 1\) - \(\frac{d}{dx} \ln (x^{2} - 2) = \frac{2x}{x^{2} - 2}\) (using the chain rule) Substitute back: \[ \frac{h'(x)}{h(x)} = 1 \cdot \ln (x^{2} - 2) + (x + 1) \cdot \frac{2x}{x^{2} - 2} \] 3. **Solve for \(h'(x)\):** Multiply both sides by \(h(x)\): \[ h'(x) = h(x) \left[ \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right] \] 4. **Substitute back \(h(x)\):** \[ h'(x) = \left( x^{2} - 2 \right)^{x + 1} \left[ \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right] \] **Simplified Form:** \[ h'(x) = \left( x^{2} - 2 \right)^{x + 1} \left[ \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right] \] --- **Summary of Derivatives:** 1. **\(f'(x)\):** \[ f'(x) = \frac{1}{2\sqrt{x}} + 3^{x} \ln 3 + \frac{6}{x^{4}} \] 2. **\(g'(x)\):** \[ g'(x) = \cos^2 x \cdot \ln x + \frac{\sin x \cos x}{x} - \sin^2 x \cdot \ln x \] *Or alternatively:* \[ g'(x) = (\cos^2 x - \sin^2 x) \cdot \ln x + \frac{\sin x \cos x}{x} \] 3. **\(h'(x)\):** \[ h'(x) = \left( x^{2} - 2 \right)^{x + 1} \left[ \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right] \] If you have any further questions or need additional clarification on any of the steps, feel free to ask!