Question

Anyone can solve this ​

Answer 1

EXPLANATION.

$\sf : \implies \: \: equation \: are \: \\ \\ \sf : \implies \: \: y {}^{ \cot(x) } + ( \tan {}^{ - 1} x ) {}^{y} = 1 \: \: then \: find \: \frac{dy}{dx}$

$\sf : \implies \: \: let \: we \: assume \: that \: \\ \\ \sf : \implies \: \: u \: = y {}^{ \cot(x) } \\ \\\sf : \implies \: \: v \: \: = (\tan {}^{ - 1} \: x) {}^{y}$

$\sf : \implies \: \: differentiate \: both \: side \: and \: add \: log \: on \: both \: sides$

$\sf : \implies \: \: u \: = \cot(x) \\ \\ \sf : \implies \: \: \frac{1}{u} . \frac{du}{dx} = \cot(x). \frac{1}{y}. \frac{dy}{dx} + log(y).( - \csc {}^{2} (x)) \\ \\ \sf : \implies \: \: \frac{du}{dx} = y {}^{ \cot(x) } \Bigg( \frac{ \cot(x) }{y} . \frac{dy}{dx} - ( \csc {}^{2} (x) log(y) \Bigg) \: \: ....(1)$

$\sf : \implies \: log(v) = y log( \tan {}^{ - 1} x ) \\ \\ \sf : \implies \: \frac{1}{v}. \frac{dv}{dx} = y. \frac{1}{ \tan {}^{ - 1} (x) }. \frac{1}{1 + {x}^{2} } + log( \tan {}^{ - 1}x). \frac{dy}{dx} \\ \\ \sf : \implies \: \frac{dv}{dx} = ( \tan {}^{ - 1} (x) \Bigg( \frac{y}{(1 + {x}^{2}) \tan {}^{ - 1} )x } + log( \tan {}^{ - 1} x ) . \frac{dy}{dx} \Bigg)...(2)$

$\sf : \implies \: from \: equation \: (1) \: \: and \: \: (2) \: \: we \: get$

$\sf : \implies \: \dfrac{du}{dx} + \dfrac{dv}{dx} = 0$

$\sf : \implies \: \dfrac{dy}{dx} = \dfrac{y {}^{ \cot \: (x) } . \csc {}^{2} (x) . log(y) - \Bigg( \dfrac{ \tan {}^{ - 1} (x) {}^{y - 1} .y}{1 + {x}^{2} } \Bigg) }{y {}^{ \cot(x) - 1 } . \cot(x) + ( \tan {}^{ - 1} (x) {}^{y}. log( \tan {}^{ - 1} x ) }$