Question:

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

Updated On: Apr 2, 2025

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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) $\frac{\ln x}{(1+\ln x)^2}$.

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List-I	List-II
The derivative of $ \log_e x $ with respect to $ \frac{1}{x} $ at $ x = 5 $ is	(I) -5
If $ x^3 + x^2y + xy^2 - 21x = 0 $, then $ \frac{dy}{dx} $ at $ (1, 1) $ is	(II) -6
If $ f(x) = x^3 \log_e \frac{1}{x} $, then $ f'(1) + f''(1) $ is	(III) 5
If $ y = f(x^2) $ and $ f'(x) = e^{\sqrt{x}} $, then $ \frac{dy}{dx} $ at $ x = 0 $ is	(IV) 0

View More Questions

View More Questions

List-I	List-II
The derivative of \( \log_e x \) with respect to \( \frac{1}{x} \) at \( x = 5 \) is	(I) -5
If \( x^3 + x^2y + xy^2 - 21x = 0 \), then \( \frac{dy}{dx} \) at \( (1, 1) \) is	(II) -6
If \( f(x) = x^3 \log_e \frac{1}{x} \), then \( f'(1) + f''(1) \) is	(III) 5
If \( y = f(x^2) \) and \( f'(x) = e^{\sqrt{x}} \), then \( \frac{dy}{dx} \) at \( x = 0 \) is	(IV) 0