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

To solve the problem, we need to find the derivative of the curve \( y = x^2 - xy \) at the point \( (1, \frac{1}{2}) \).

1. Differentiate the Given Equation:
We start by differentiating both sides of \( y = x^2 - xy \) with respect to \( x \):

\( \frac{dy}{dx} = \frac{d}{dx}(x^2) - \frac{d}{dx}(xy) \).

2. Apply Product Rule to \( xy \):
Using the product rule for \( \frac{d}{dx}(xy) \), we get:

\( \frac{dy}{dx} = 2x - \left( x \frac{dy}{dx} + y \cdot 1 \right) \).

3. Collect Like Terms:
Gather all \( \frac{dy}{dx} \) terms on one side:

\( \frac{dy}{dx} + x \frac{dy}{dx} = 2x - y \).

4. Factor Out \( \frac{dy}{dx} \):
Factor the left-hand side to solve for \( \frac{dy}{dx} \):

\( (1 + x) \frac{dy}{dx} = 2x - y \).

5. Solve for the Derivative:
Isolate \( \frac{dy}{dx} \) to find the general expression for the derivative:

\( \frac{dy}{dx} = \frac{2x - y}{1 + x} \).

6. Evaluate at the Given Point \( (1, \frac{1}{2}) \):
Substitute \( x = 1 \) and \( y = \frac{1}{2} \) into the derivative:

\( \left. \frac{dy}{dx} \right|_{(1, \frac{1}{2})} = \frac{2(1) - \frac{1}{2}}{1 + 1} = \frac{\frac{3}{2}}{2} = \frac{3}{4} \).

Final Answer:
The derivative of the curve at the point \( (1, \frac{1}{2}) \) is \( \boxed{\dfrac{3}{4}} \).

Answer 2

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 \( \frac{3}{4} \).