Question

If $x+y=3e^2$ then $\frac{d}{dx}\left(x^y\right)=0$ for $x=$.

• A. e
• B. $e^2$
• C. $e^e$
• D. $2e^2$

Answer 1

The correct option is B $e^2$

Let $x^y=t$

taking log on both sides, we get

$y\ln x=\ln t$

Differentiating both sides w.r.t. $x$ we get

$\frac{y}{x}+\ln x.\frac{dy}{dx}=\frac{1}{t}.\frac{dt}{dx}$

$\Rightarrow t(\frac{y}{x}+\ln x.\frac{dy}{dx})=\frac{dt}{dx}.............\left(1\right)$

From $x+y=3e^2$

$\Rightarrow \frac{dy}{dx}=-1$

Putting it in (1) we get

$\Rightarrow t(\frac{y}{x}-\ln x)=\frac{dt}{dx}=0$

So $\frac{t}{x}=0$ or $y-x\ln x=0$

$\Rightarrow y-x\ln x=0$ as $\frac{t}{x}$is non-zero for given values of $x$.

Since $x+y=3e^2$

$\Rightarrow x(1+\ln x)=3e^2..............\left(2\right)$

Out of given options $x=e^2$ satisfies (2)

Hence option B is correct.