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# Solution for If X Y = E X − Y , Then D Y D X is (A) 1 + X 1 + Log X - CBSE (Commerce) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

#### Question

If $x^y = e^{x - y} ,\text{ then } \frac{dy}{dx}$ is

(a) $\frac{1 + x}{1 + \log x}$

(b) $\frac{1 - \log x}{1 + \log x}$

(c) not defined

(d) $\frac{\log x}{\left( 1 + \log x \right)^2}$

#### Solution

(d)

$\frac{\log x}{\left( 1 + \log x \right)^2}$

$\text{ We have,} x^y = e^{x - y}$
$\text{ Taking log on both sides we get },$
$\Rightarrow y \log x = \left( x - y \right) \log_e e$
$\Rightarrow y \log x = x - y$
$\Rightarrow y \log x + y = x$
$\Rightarrow y\left( 1 + \log x \right) = x$
$\Rightarrow y = \frac{x}{1 + \log x}$

$\Rightarrow \frac{dy}{dx} = \frac{\left( 1 + \log x \right) \times 1 - x \times \left( 0 + \frac{1}{x} \right)}{\left( 1 + \log x \right)^2}$
$\Rightarrow \frac{dy}{dx} = \frac{1 + \log x - 1}{\left( 1 + \log x \right)^2}$
$\Rightarrow \frac{dy}{dx} = \frac{\log x}{\left( 1 + \log x \right)^2}$

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Solution If X Y = E X − Y , Then D Y D X is (A) 1 + X 1 + Log X Concept: Simple Problems on Applications of Derivatives.
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