8 (i) Given that $$ \frac{2 x-1}{(x-1)^{2}}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}} $$ find the value of $$ A $$ and of $$ B $$. (ii) Hence find the derivative of $$ \frac{2 x-1}{(x-1)^{2}} $$.

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Certainly! Let's tackle each part of the problem step by step.

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### **Problem 8 (i)**

Given:
$$
\frac{2x - 1}{(x - 1)^2} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2}
$$
We need to find the values of $$ A $$ and $$ B $$.

#### **Step 1: Combine the Right-Hand Side (RHS) Fractions**

To combine the fractions on the RHS, express them with a common denominator:
$$
\frac{A}{x - 1} + \frac{B}{(x - 1)^2} = \frac{A(x - 1) + B}{(x - 1)^2}
$$

#### **Step 2: Equate the Numerators**

Since the denominators are the same, equate the numerators:
$$
A(x - 1) + B = 2x - 1
$$

#### **Step 3: Expand and Simplify**

Expand the left side:
$$
A \cdot x - A + B = 2x - 1
$$
$$
Ax + (-A + B) = 2x - 1
$$

#### **Step 4: Match the Coefficients**

For the equation to hold for all $$ x $$, the coefficients of corresponding powers of $$ x $$ must be equal:

1. **Coefficient of $$ x $$:**
   $$
   A = 2
   $$

2. **Constant Term:**
   $$
   -A + B = -1
   $$
   Substitute $$ A = 2 $$:
   $$
   -2 + B = -1 \quad \Rightarrow \quad B = 1
   $$

#### **Conclusion for Part (i):**
$$
A = 2 \quad \text{and} \quad B = 1
$$

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### **Problem 8 (ii)**

Using the values of $$ A $$ and $$ B $$ found in part (i), we now need to find the derivative of:
$$
f(x) = \frac{2x - 1}{(x - 1)^2}
$$

Given the partial fraction decomposition:
$$
f(x) = \frac{2}{x - 1} + \frac{1}{(x - 1)^2}
$$

#### **Method 1: Differentiating Term by Term**

Differentiate each term separately.

1. **Differentiate $$ \frac{2}{x - 1} $$:**
   $$
   \frac{d}{dx} \left( \frac{2}{x - 1} \right) = 2 \cdot \frac{d}{dx} \left( (x - 1)^{-1} \right) = 2 \cdot (-1)(x - 1)^{-2} = -\frac{2}{(x - 1)^2}
   $$

2. **Differentiate $$ \frac{1}{(x - 1)^2} $$:**
   $$
   \frac{d}{dx} \left( \frac{1}{(x - 1)^2} \right) = \frac{d}{dx} \left( (x - 1)^{-2} \right) = -2(x - 1)^{-3} = -\frac{2}{(x - 1)^3}
   $$

3. **Combine the Derivatives:**
   $$
   f'(x) = -\frac{2}{(x - 1)^2} - \frac{2}{(x - 1)^3}
   $$
   $$
   f'(x) = -\frac{2(x - 1) + 2}{(x - 1)^3} = -\frac{2x - 2 + 2}{(x - 1)^3} = -\frac{2x}{(x - 1)^3}
   $$

#### **Method 2: Using the Quotient Rule**

Alternatively, differentiate $$ f(x) $$ directly using the quotient rule:
$$
f(x) = \frac{2x - 1}{(x - 1)^2}
$$
$$
f'(x) = \frac{d}{dx}\left(2x - 1\right) \cdot (x - 1)^2 - (2x - 1) \cdot \frac{d}{dx}\left((x - 1)^2\right)}{(x - 1)^4}
$$
$$
f'(x) = \frac{2(x - 1)^2 - (2x - 1)(2)(x - 1)}{(x - 1)^4}
$$
$$
f'(x) = \frac{2(x^2 - 2x + 1) - 2(2x - 1)(x - 1)}{(x - 1)^4}
$$
$$
f'(x) = \frac{2x^2 - 4x + 2 - (4x^2 - 6x + 2)}{(x - 1)^4}
$$
$$
f'(x) = \frac{-2x^2 + 2x}{(x - 1)^4} = -\frac{2x(x - 1)}{(x - 1)^4} = -\frac{2x}{(x - 1)^3}
$$

Both methods yield the same result.

#### **Conclusion for Part (ii):**
$$
f'(x) = -\frac{2x}{(x - 1)^3} \quad \text{or equivalently} \quad f'(x) = -\frac{2}{(x - 1)^2} - \frac{2}{(x - 1)^3}
$$

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### **Final Answer:**

(i) $$ A = 2 $$ and $$ B = 1 $$.

(ii) The derivative is $$ f\,'(x) = -\dfrac{2x}{\, (x - 1)^3\,} $$.
$$ A = 2 $$ and $$ B = 1 $$. The derivative is $$ f\,'(x) = -\dfrac{2x}{(x - 1)^3} $$.

8 (i) Given that \( \frac{2 x-1}{(x-1)^{2}}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}} \) find the value of \( A \) and of \( B \). (ii) Hence find the derivative of \( \frac{2 x-1}{(x-1)^{2}} \).

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