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If $y=\log _y x$, then $\frac{d y}{d x}$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{x(1+\log y)}$
$\because y=\log _y x$
To find, $\frac{d y}{d x}$
$\Rightarrow \quad y=\frac{\log _e x}{\log _e y}$
$y \log y=\log x$
On differentiating w.r.t. $x$,
$y \cdot \frac{d}{d x} \log y+\log y \cdot \frac{d y}{d x}=\frac{d}{d x} \log x$
$\Rightarrow \quad y \cdot \frac{1}{y} \cdot \frac{d y}{d x}+\log y \frac{d y}{d x}=\frac{1}{x}$
$\Rightarrow \quad(1+\log y) \frac{d y}{d x}=\frac{1}{x}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{1}{x(1+\log y)}$
To find, $\frac{d y}{d x}$
$\Rightarrow \quad y=\frac{\log _e x}{\log _e y}$
$y \log y=\log x$
On differentiating w.r.t. $x$,
$y \cdot \frac{d}{d x} \log y+\log y \cdot \frac{d y}{d x}=\frac{d}{d x} \log x$
$\Rightarrow \quad y \cdot \frac{1}{y} \cdot \frac{d y}{d x}+\log y \frac{d y}{d x}=\frac{1}{x}$
$\Rightarrow \quad(1+\log y) \frac{d y}{d x}=\frac{1}{x}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{1}{x(1+\log y)}$
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