Q100 / 160 · TS EAMCET 2006 · Mathematics

If $x^y=y^x$, then $x(x-y \log x) \frac{d y}{d x}$ is equal to :

  • Correct answer
    A
    $y(y-x \log y)$
  • B
    $y(y+x \log y)$
  • C
    $x(x+y \log x)$
  • D
    $x(y-x \log y)$

Solution

$\because x^y=y^x$ Taking log on both sides, we get $y \log x=x \log y$ On differentiating with respect to $x$, we get $y \cdot \frac{1}{x}+\log x \frac{d y}{d x}=x \cdot \frac{1}{y} \frac{d y}{d x}+\log y$ $\Rightarrow \quad \frac{(x-y \log x)}{y} \frac{d y}{d x}=\frac{-x \log y+y}{x}$ $\Rightarrow \quad x(x-y \log x) \frac{d y}{d x}=y(-x \log y+y)$

Hence, the correct answer is option: 'A'.

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