Question

The slope of the tangent to the curve, $ y = x^2 - xy $ at $ (1, \frac{1}{2}) $ is

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

Hint:

When differentiating implicitly, remember to apply the product rule when differentiating terms like \( xy \).

Answer 1

The Correct Option is C

We are given the equation of the curve \( y = x^2 - xy \).
To find the slope of the tangent, we need to compute the derivative \( \frac{dy}{dx} \) and then evaluate it at the point \( (1, \frac{1}{2}) \). 
First, differentiate the equation implicitly with respect to \( x \): \[ \frac{d}{dx}(y) = \frac{d}{dx}(x^2 - xy) \] Using the product rule for \( xy \): \[ \frac{dy}{dx} = 2x - \left( x \frac{dy}{dx} + y \right) \] Now, substitute \( x = 1 \) and \( y = \frac{1}{2} \) into the equation: \[ \frac{dy}{dx} = 2(1) - \left( 1 \cdot \frac{dy}{dx} + \frac{1}{2} \right) \] \[ \frac{dy}{dx} = 2 - \left( \frac{dy}{dx} + \frac{1}{2} \right) \] \[ \frac{dy}{dx} + \frac{dy}{dx} = 2 - \frac{1}{2} \] \[ 2 \frac{dy}{dx} = \frac{3}{2} \] \[ \frac{dy}{dx} = \frac{3}{4} \] Thus, the slope of the tangent at the point \( (1, \frac{1}{2}) \) is \( \frac{3}{4} \).