Question

If \( y = \frac{x^2}{x - 1} \), then \( \frac{dy}{dx} \) at \( x = -1 \) is:

• A. \( \frac{1}{4} \)
• B. \( -\frac{1}{4} \)
• C. 1
• D. \( -\frac{1}{2} \)
• E. \( \frac{3}{4} \)

Hint:

Use the quotient rule to differentiate rational functions: \( \frac{dy}{dx} = \frac{v \cdot u' - u \cdot v'}{v^2} \), where \( u \) and \( v \) are the numerator and denominator, respectively.

Answer 1

Answer: (E)

We are given \( y = \frac{x^2}{x - 1} \). 
To differentiate \( y \), we use the quotient rule: \[ \frac{dy}{dx} = \frac{(x - 1)(2x) - x^2(1)}{(x - 1)^2} \] Simplifying the numerator: \[ = 2x(x - 1) - x^2 = 2x^2 - 2x - x^2 = x^2 - 2x \] Thus: \[ \frac{dy}{dx} = \frac{x^2 - 2x}{(x - 1)^2} \] Now, substitute \( x = -1 \): \[ \frac{dy}{dx} = \frac{(-1)^2 - 2(-1)}{((-1) - 1)^2} = \frac{1 + 2}{(-2)^2} = \frac{3}{4} \] Thus, \( \frac{dy}{dx} = \frac{3}{4} \) at \( x = -1 \).