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: Simplify the given equation.
We are given:
\[ (xe)^y-e^x=0 \] So, \[ (xe)^y=e^x \]

Step 2: Take natural logarithm on both sides.
Applying \(\ln\) on both sides, we get:
\[ \ln\left((xe)^y\right)=\ln(e^x) \] Using logarithmic laws:
\[ y\ln(xe)=x \]

Step 3: Simplify \(\ln(xe)\).
Since \[ \ln(xe)=\ln x+\ln e=\ln x+1, \] the equation becomes:
\[ y(\ln x+1)=x \]

Step 4: Differentiate implicitly with respect to \(x\).
Differentiate both sides:
\[ \frac{d}{dx}\big(y(\ln x+1)\big)=\frac{d}{dx}(x) \] Using the product rule:
\[ \frac{dy}{dx}(\ln x+1)+y\left(\frac{1}{x}\right)=1 \]

Step 5: Find the value of \(y\) at \(x=1\).
Substitute \(x=1\) into \[ y(\ln x+1)=x \] to get:
\[ y(\ln 1+1)=1 \] \[ y(0+1)=1 \] \[ y=1 \]

Step 6: Substitute \(x=1\) and \(y=1\) into the derivative equation.
From \[ \frac{dy}{dx}(\ln x+1)+\frac{y}{x}=1, \] put \(x=1\) and \(y=1\):
\[ \frac{dy}{dx}(0+1)+\frac{1}{1}=1 \] \[ \frac{dy}{dx}+1=1 \] \[ \frac{dy}{dx}=0 \]

Step 7: State the final answer.
Hence, \[ \boxed{0} \] which matches option \((1)\).