Q105 / 180 · COMEDK 2020 · Mathematics

If $x^{x}=y^{y}$, then $\frac{d y}{d x}$ is

  • A
    $-\frac{x}{y}$
  • B
    $-\frac{y}{x}$
  • Correct answer
    C
    $\frac{1+\log x}{1+\log y}$
  • D
    $1+\log \left(\frac{x}{y}\right)$

Solution

We have, $x^{x}=y^{y} \Rightarrow x \log x=y \log y$ $\Rightarrow \quad 1 \cdot \log x+\frac{x}{x}=y^{\prime} \log y+\frac{y}{y} y^{\prime}$ $\Rightarrow \quad \log x+1=y^{\prime}(\log y+1)$ $\Rightarrow \quad \frac{d y}{d x}=\frac{1+\log x}{1+\log y}$

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

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