Question

If \( (x^2-3x+2)e^{\frac{y}{x-1}} = x+2 \) then \( \left(\frac{dy}{dx}\right)_{x=0} = \)

• A. 2
• B. -2
• C. 1
• D. -1

Hint:

When finding a derivative at a specific point for an implicit function, substitute the coordinate values immediately after differentiating. This simplifies the algebra significantly compared to solving for the general \( dy/dx \) first.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

This is an implicit differentiation problem. We need to find the derivative \( \frac{dy}{dx} \) at a specific point \( x=0 \). First, we must find the corresponding \( y \) value at \( x=0 \). It is often easier to simplify the equation using logarithms before differentiating.
Step 2: Key Formula or Approach:

1. Logarithmic Differentiation: \( \ln(A \cdot B) = \ln A + \ln B \). 2. Product and Quotient rules of differentiation.
Step 3: Detailed Explanation:

Given equation: \[ (x^2 - 3x + 2) e^{\frac{y}{x-1}} = x + 2 \] Factorize the quadratic term: \( x^2 - 3x + 2 = (x-1)(x-2) \). \[ (x-1)(x-2) e^{\frac{y}{x-1}} = x + 2 \] Find \( y \) when \( x = 0 \): Substitute \( x=0 \) into the original equation: \[ (0^2 - 0 + 2) e^{\frac{y}{0-1}} = 0 + 2 \] \[ 2 e^{-y} = 2 \] \[ e^{-y} = 1 \implies -y = 0 \implies y = 0 \] So we need to find \( \frac{dy}{dx} \) at \( (0, 0) \). Take the natural logarithm (ln) on both sides of the factored equation: \[ \ln\left( (x-1)(x-2) e^{\frac{y}{x-1}} \right) = \ln(x+2) \] \[ \ln(x-1) + \ln(x-2) + \frac{y}{x-1} = \ln(x+2) \] Note: Since we are evaluating at \( x=0 \), the arguments of logs like \( x-1 \) would be negative. However, the original equation is valid. The logarithm of products actually uses absolute values or we can differentiate the original form directly to avoid domain issues with complex logs. Let's differentiate the original implicit form or the rearranged form. Rearranged form: \[ \frac{y}{x-1} = \ln(x+2) - \ln(x^2 - 3x + 2) \] Differentiating with respect to \( x \): \[ \frac{(x-1)\frac{dy}{dx} - y(1)}{(x-1)^2} = \frac{1}{x+2} - \frac{2x - 3}{x^2 - 3x + 2} \] Substitute \( x = 0 \) and \( y = 0 \): \[ \frac{(0-1)y' - 0}{(0-1)^2} = \frac{1}{0+2} - \frac{2(0) - 3}{0 - 0 + 2} \] \[ \frac{-y'}{1} = \frac{1}{2} - \frac{-3}{2} \] \[ -y' = \frac{1}{2} + \frac{3}{2} \] \[ -y' = 2 \] \[ y' = -2 \]
Step 4: Final Answer:

The value of \( \frac{dy}{dx} \) at \( x=0 \) is -2.