Question

pls answer question​

Answer 1

$\huge \: \red \star \: { \large \pink { \underline\mathcal{ANSWER}} } \: \red\star \: \:$

option (d)

$\green{standered \: results}$

• $\footnotesize{ log_{ \: a}(a) = 1}$
• $\footnotesize { log \: {x}^{n} = \: n \: log \: x }$

$\green{ \small {solution}} \:$

$\mathrm{given} \\ \small{ {x}^{y} = {e}^{x + y} }$

on taking log on both sides

$\small{y \: log(x) = (x + y) log_{e}(e) }$

now differentiate wrt x

$\small{ \frac{dy}{dx} \: logx \: + y \: ( \frac{1}{x} )} \: = \small{\: 1 + \frac{dy}{dx} } \\ \\ \small{\frac{dy}{dx} \: log x } \: - \small{ \frac{dy}{dx} \: = 1 - \frac{x}{y} }$

$\frac{dy}{dx} \: \small{ ( log(x) - 1)} = \small{ \frac{x - y}{x} }$

$\frac{dy}{dx} = \frac{x - y}{x \: \footnotesize{( log(x) - 1)} }$

hope it helps

Answer 2

Step-by-step explanation:

Option D is ur correct answer ❤️❤️