(i) If `y= (x-1)/( 2x^(2) - 7x +5)`, find `(dy)/( dx)` at `x=2`.

Answer

To find \(\frac{dy}{dx}\) for the function \(y = \frac{x-1}{2x^2 - 7x + 5}\) at \(x = 2\), we will use the quotient rule for differentiation. The quotient rule states that if \(y = \frac{u}{v}\), then \(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\).

### Step 1: Identify \(u\) and \(v\)
Let:
- \(u = x - 1\)
- \(v = 2x^2 - 7x + 5\)

### Step 2: Differentiate \(u\) and \(v\)
Now, we need to find \(\frac{du}{dx}\) and \(\frac{dv}{dx}\).

1. Differentiate \(u\):
\[ \frac{du}{dx} = 1 \]

2. Differentiate \(v\):
\[ \frac{dv}{dx} = \frac{d}{dx}(2x^2 - 7x + 5) = 4x - 7 \]

### Step 3: Apply the Quotient Rule
Using the quotient rule:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Substituting \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\):
\[ \frac{dy}{dx} = \frac{(2x^2 - 7x + 5)(1) - (x - 1)(4x - 7)}{(2x^2 - 7x + 5)^2} \]

### Step 4: Simplify the Expression
Now we simplify the numerator:
\[ = (2x^2 - 7x + 5) - (x - 1)(4x - 7) \]
Expanding \((x - 1)(4x - 7)\):
\[ = 4x^2 - 7x - 4x + 7 = 4x^2 - 11x + 7 \]
Thus, the numerator becomes:
\[ 2x^2 - 7x + 5 - (4x^2 - 11x + 7) = 2x^2 - 7x + 5 - 4x^2 + 11x - 7 \]
Combining like terms:
\[ = -2x^2 + 4x - 2 \]

### Step 5: Final Expression for \(\frac{dy}{dx}\)
Now we have:
\[ \frac{dy}{dx} = \frac{-2x^2 + 4x - 2}{(2x^2 - 7x + 5)^2} \]

### Step 6: Evaluate at \(x = 2\)
Now we substitute \(x = 2\):
1. Calculate the numerator:
\[ -2(2^2) + 4(2) - 2 = -8 + 8 - 2 = -2 \]

2. Calculate the denominator:
\[ 2(2^2) - 7(2) + 5 = 8 - 14 + 5 = -1 \]
Therefore,
\[ (2x^2 - 7x + 5)^2 = (-1)^2 = 1 \]

Putting it all together:
\[ \frac{dy}{dx} \bigg|_{x=2} = \frac{-2}{1} = -2 \]

### Final Answer
Thus, \(\frac{dy}{dx}\) at \(x = 2\) is \(-2\).

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