1 Differentiate the following functions with respect to \( x \). (a) \( x(3 x+2)^{2} \) (b) \( (x+1)(x-3)^{2} \) (c) \( (2-3 x)(4 x+1)^{4} \) (d) \( (2+x)^{2}\left(x^{2}+2\right) \)

Certainly! Let's differentiate each of the given functions with respect to \( x \). We'll use differentiation rules such as the **Product Rule** and the **Chain Rule** where applicable. ### Problem 1(a): **Differentiate** \( f(x) = x(3x + 2)^2 \) **with respect to** \( x \). **Solution:** We'll use the **Product Rule**, which states that if \( f(x) = u(x) \cdot v(x) \), then \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \] Let: \[ u(x) = x \quad \text{and} \quad v(x) = (3x + 2)^2 \] First, find \( u'(x) \) and \( v'(x) \): 1. **Differentiate \( u(x) = x \):** \[ u'(x) = 1 \] 2. **Differentiate \( v(x) = (3x + 2)^2 \) using the **Chain Rule**: \[ v'(x) = 2(3x + 2) \cdot \frac{d}{dx}(3x + 2) = 2(3x + 2) \cdot 3 = 6(3x + 2) \] Now, apply the Product Rule: \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = 1 \cdot (3x + 2)^2 + x \cdot 6(3x + 2) \] Simplify the expression: \[ f'(x) = (3x + 2)^2 + 6x(3x + 2) \] \[ f'(x) = (3x + 2)\left[(3x + 2) + 6x\right] = (3x + 2)(9x + 2) \] **Final Answer:** \[ f'(x) = (3x + 2)(9x + 2) \] --- ### Problem 1(b): **Differentiate** \( f(x) = (x + 1)(x - 3)^2 \) **with respect to** \( x \). **Solution:** Again, we'll use the **Product Rule**. Let: \[ u(x) = x + 1 \quad \text{and} \quad v(x) = (x - 3)^2 \] First, find \( u'(x) \) and \( v'(x) \): 1. **Differentiate \( u(x) = x + 1 \):** \[ u'(x) = 1 \] 2. **Differentiate \( v(x) = (x - 3)^2 \) using the **Chain Rule**: \[ v'(x) = 2(x - 3) \cdot \frac{d}{dx}(x - 3) = 2(x - 3) \cdot 1 = 2(x - 3) \] Apply the Product Rule: \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = 1 \cdot (x - 3)^2 + (x + 1) \cdot 2(x - 3) \] Factor out \( (x - 3) \): \[ f'(x) = (x - 3)\left[(x - 3) + 2(x + 1)\right] \] \[ f'(x) = (x - 3)(x - 3 + 2x + 2) = (x - 3)(3x - 1) \] **Final Answer:** \[ f'(x) = (x - 3)(3x - 1) \] --- ### Problem 1(c): **Differentiate** \( f(x) = (2 - 3x)(4x + 1)^4 \) **with respect to** \( x \). **Solution:** We'll employ the **Product Rule** once more. Let: \[ u(x) = 2 - 3x \quad \text{and} \quad v(x) = (4x + 1)^4 \] First, find \( u'(x) \) and \( v'(x) \): 1. **Differentiate \( u(x) = 2 - 3x \):** \[ u'(x) = -3 \] 2. **Differentiate \( v(x) = (4x + 1)^4 \) using the **Chain Rule**: \[ v'(x) = 4(4x + 1)^3 \cdot \frac{d}{dx}(4x + 1) = 4(4x + 1)^3 \cdot 4 = 16(4x + 1)^3 \] Apply the Product Rule: \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = (-3)(4x + 1)^4 + (2 - 3x) \cdot 16(4x + 1)^3 \] Factor out \( (4x + 1)^3 \): \[ f'(x) = (4x + 1)^3 \left[ -3(4x + 1) + 16(2 - 3x) \right] \] \[ = (4x + 1)^3 \left[ -12x - 3 + 32 - 48x \right] \] \[ = (4x + 1)^3 (-60x + 29) \] **Final Answer:** \[ f'(x) = (4x + 1)^3 (29 - 60x) \] --- ### Problem 1(d): **Differentiate** \( f(x) = (2 + x)^2 (x^2 + 2) \) **with respect to** \( x \). **Solution:** We'll use the **Product Rule**. Let: \[ u(x) = (2 + x)^2 \quad \text{and} \quad v(x) = x^2 + 2 \] First, find \( u'(x) \) and \( v'(x) \): 1. **Differentiate \( u(x) = (2 + x)^2 \) using the **Chain Rule**: \[ u'(x) = 2(2 + x) \cdot \frac{d}{dx}(2 + x) = 2(2 + x) \cdot 1 = 2(2 + x) \] 2. **Differentiate \( v(x) = x^2 + 2 \):** \[ v'(x) = 2x \] Apply the Product Rule: \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = 2(2 + x)(x^2 + 2) + (2 + x)^2 \cdot 2x \] Factor out \( 2(2 + x) \): \[ f'(x) = 2(2 + x)\left[ x^2 + 2 + x(2 + x) \right] \] \[ = 2(2 + x)\left[ x^2 + 2 + 2x + x^2 \right] \] \[ = 2(2 + x)(2x^2 + 2x + 2) \] We can factor out a 2 from the quadratic polynomial: \[ = 2(2 + x) \cdot 2(x^2 + x + 1) = 4(2 + x)(x^2 + x + 1) \] **Final Answer:** \[ f'(x) = 4(x + 2)(x^2 + x + 1) \] --- **Summary of Derivatives:** 1. **(a)** \( \displaystyle f'(x) = (3x + 2)(9x + 2) \) 2. **(b)** \( \displaystyle f'(x) = (x - 3)(3x - 1) \) 3. **(c)** \( \displaystyle f'(x) = (4x + 1)^3 (29 - 60x) \) 4. **(d)** \( \displaystyle f'(x) = 4(x + 2)(x^2 + x + 1) \) If you have any further questions or need additional clarification on any of the steps, feel free to ask!