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 $x = -1$.