Question

If $(xe)^y-e^x=0$, then $\frac{dy}{dx}$ at $x=1$ is:

• A. \(0\)
• B. \(2\)
• C. \(\frac{1}{2}\)
• D. \(\frac{1}{4}\)
• E. \(4\)

Hint:

In implicit differentiation, first simplify the equation as much as possible. Logarithms are very useful when variables appear in exponents.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The given equation defines \( y \) implicitly as a function of \( x \). To find \( \frac{dy}{dx} \), we use implicit differentiation. A helpful first step is to simplify the equation using properties of logarithms to make differentiation easier.
Step 2: Key Formula or Approach:
1. Simplify the equation: \( (xe)^y = e^x \).
2. Take the natural logarithm of both sides to bring the exponent \( y \) down.
3. Differentiate the resulting equation implicitly with respect to \( x \), remembering to treat \( y \) as a function of \( x \) and applying the chain rule.
4. Solve for \( \frac{dy}{dx} \).
5. Substitute \( x=1 \) to find the value of \( y \) and then the value of \( \frac{dy}{dx} \).
Step 3: Detailed Explanation:
The equation is \( (xe)^y = e^x \).
Take the natural logarithm of both sides:
\[ \ln((xe)^y) = \ln(e^x) \] Using the logarithm property \( \ln(a^b) = b \ln(a) \):
\[ y \ln(xe) = x \] Using the property \( \ln(ab) = \ln(a) + \ln(b) \):
\[ y(\ln x + \ln e) = x \] Since \( \ln e = 1 \):
\[ y(\ln x + 1) = x \] Now, differentiate both sides with respect to \( x \). We use the product rule on the left side:
\[ \frac{d}{dx}[y(\ln x + 1)] = \frac{d}{dx}[x] \] \[ \left(\frac{dy}{dx}\right)(\ln x + 1) + y \cdot \frac{d}{dx}(\ln x + 1) = 1 \] \[ \left(\frac{dy}{dx}\right)(\ln x + 1) + y \cdot \left(\frac{1}{x}\right) = 1 \] We need to find the value of \( y \) when \( x=1 \). Substitute \( x=1 \) into the simplified equation \( y(\ln x + 1) = x \):
\[ y(\ln 1 + 1) = 1 \] \[ y(0 + 1) = 1 \implies y = 1 \] Now substitute \( x=1 \) and \( y=1 \) into the differentiated equation:
\[ \left(\frac{dy}{dx}\right)(\ln 1 + 1) + (1) \cdot \left(\frac{1}{1}\right) = 1 \] \[ \left(\frac{dy}{dx}\right)(0 + 1) + 1 = 1 \] \[ \frac{dy}{dx} + 1 = 1 \] \[ \frac{dy}{dx} = 0 \] Step 4: Final Answer:
The value of \( \frac{dy}{dx} \) at \( x=1 \) is 0. This corresponds to option (A).