Question

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

• A. \(\frac{\log x}{(1+\log x)^2}\)
• B. \(\frac{1}{(1+\log x)^2}\)
• C. \(\frac{\log x}{(1+\log x)}\)
• D. \(\frac{e^x}{x(y-1)}\)

Answer 1

The Correct Option is A

If \((xe)^y = e^x\), then we need to find \(\frac{dy}{dx}\).

Take the natural logarithm of both sides:

\(\ln((xe)^y) = \ln(e^x)\)

\(y \ln(xe) = x\)

\(y (\ln x + \ln e) = x\)

\(y (\ln x + 1) = x\)

\(y = \frac{x}{1+\ln x}\)

Now, we differentiate with respect to x:

\(\frac{dy}{dx} = \frac{d}{dx}(\frac{x}{1+\ln x})\)

Using the quotient rule: \(\frac{d}{dx} \frac{u}{v} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)

\(\frac{dy}{dx} = \frac{(1+\ln x)(1) - x(\frac{1}{x})}{(1+\ln x)^2}\)

\(\frac{dy}{dx} = \frac{1+\ln x - 1}{(1+\ln x)^2}\)

\(\frac{dy}{dx} = \frac{\ln x}{(1+\ln x)^2}\)

Therefore, the correct option is (A) \(\frac{\ln x}{(1+\ln x)^2}\).

Answer 2

Given $(xe)^y = e^x$, taking the natural logarithm of both sides, 

we have: $y \ln(xe) = x$ $y(\ln x + \ln e) = x$ $y(\ln x + 1) = x$ $y = \frac{x}{1+\ln x}$ 

Differentiating with respect to $x$, we have: $\frac{dy}{dx} = \frac{(1+\ln x)(1) - x(\frac{1}{x})}{(1+\ln x)^2} = \frac{1 + \ln x - 1}{(1+\ln x)^2} = \frac{\ln x}{(1+\ln x)^2}$