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If $x^y=y^{\sin x}(\tan x)^{\cos x}$, then $\left(\log x-\frac{\sin x}{y}\right) \frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$\cos x \log y-\sin x \log (\tan x)+\operatorname{cosec} x-\frac{y}{x}$
We have,
$x^y=y^{\sin x}(\tan x)^{\cos x}$
$\begin{aligned} & \Rightarrow \quad y \log x=\sin x \log y+\cos x \log \tan x \\ & \Rightarrow \frac{d y}{d x} \log x+\frac{y}{x}=\cos x \log y+\frac{\sin x}{y} \frac{d y}{d x} \\ & \quad-\sin x \log \tan x+\cos x \frac{1}{\tan x} \sec ^2 x \\ & \begin{array}{r}\Rightarrow\left(\log x-\frac{\sin x}{y}\right) \frac{d y}{d x}=\cos x \log y-\sin x \log \tan x \\ +\operatorname{cosec} x-\frac{y}{x}\end{array}\end{aligned}$
$x^y=y^{\sin x}(\tan x)^{\cos x}$
$\begin{aligned} & \Rightarrow \quad y \log x=\sin x \log y+\cos x \log \tan x \\ & \Rightarrow \frac{d y}{d x} \log x+\frac{y}{x}=\cos x \log y+\frac{\sin x}{y} \frac{d y}{d x} \\ & \quad-\sin x \log \tan x+\cos x \frac{1}{\tan x} \sec ^2 x \\ & \begin{array}{r}\Rightarrow\left(\log x-\frac{\sin x}{y}\right) \frac{d y}{d x}=\cos x \log y-\sin x \log \tan x \\ +\operatorname{cosec} x-\frac{y}{x}\end{array}\end{aligned}$
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