Question

If xy = e^(x-y), then the value of \(\frac{dy}{dx}\) at (1, 1) is :

• A. \(\frac{1}{2}\)
• B. 0
• C. \(\frac{1}{8}\)
• D. 1

Answer 1

The Correct Option is B

Given the equation \(xy = e^{(x-y)}\), we want to find \(\frac{dy}{dx}\) at the point (1, 1). Start by differentiating both sides of the equation with respect to \(x\). On the left, use the product rule: \(\frac{d}{dx}(xy) = x\frac{dy}{dx} + y\). On the right, differentiate \(e^{(x-y)}\) using the chain rule: \(\frac{d}{dx}(e^{(x-y)}) = e^{(x-y)}\left(1 - \frac{dy}{dx}\right)\). Thus, the differentiated equation is:

\[ x\frac{dy}{dx} + y = e^{(x-y)}\left(1 - \frac{dy}{dx}\right) \]

Substitute \(x = 1\) and \(y = 1\) into the equation:

\[ 1\cdot\frac{dy}{dx} + 1 = e^{(1-1)}\left(1 - \frac{dy}{dx}\right) \]

Simplify to obtain:

\[ \frac{dy}{dx} + 1 = 1 - \frac{dy}{dx} \]

Combine similar terms:

\[ 2\frac{dy}{dx} = 0 \]

Thus, \(\frac{dy}{dx} = 0\).

Therefore, the value of \(\frac{dy}{dx}\) at (1, 1) is 0.