Question

If \(e^{-x} + e^{-y} = 2\), then \(\frac{dy}{dx}\) is

• A. \(e^{x-y}\)
• B. \(e^{y-x}\)
• C. \(-e^{x-y}\)
• D. \(-e^{y-x}\)

Hint:

Differentiate both sides with respect to x using the chain rule on e^{-y}, then solve for dy/dx.

Answer 1

The Correct Option is D

Step 1: Understand the problem:
The relation ties \(x\) and \(y\) together without solving for \(y\).
So we differentiate both sides with respect to \(x\) and treat \(y\) as a function of \(x\).

Step 2: Differentiate each term:
The derivative of \(e^{-x}\) is \(-e^{-x}\).
By the chain rule, the derivative of \(e^{-y}\) is \(-e^{-y}\dfrac{dy}{dx}\).
The derivative of the constant 2 is 0.
\[ -e^{-x} - e^{-y}\frac{dy}{dx} = 0 \]

Step 3: Solve for dy/dx:
Move the first term to the right side.
\[ -e^{-y}\frac{dy}{dx} = e^{-x} \]
Divide both sides by \(-e^{-y}\).
\[ \frac{dy}{dx} = -\frac{e^{-x}}{e^{-y}} = -e^{-x+y} = -e^{y-x} \]

Step 4: Compare with the options:
Option (A) \(e^{x-y}\) has the wrong sign and the wrong exponent, so it is wrong.
Option (B) \(e^{y-x}\) has the right exponent but the sign is positive, so it is wrong.
Option (C) \(-e^{x-y}\) has the right sign but the exponent is reversed, so it is wrong.
Option (D) \(-e^{y-x}\) matches our result exactly.

Final Answer:
The derivative is \(-e^{y-x}\), which is option (D). \[ \boxed{\frac{dy}{dx} = -e^{y-x}} \]