Question

differentiate the following with respect to x tan-1{(1-x²)/(1+x²)}​

Answer 1

Step-by-step explanation:

We have,

$\sf \: y = tan^{ - 1} \bigg( \frac{1 - {x}^{2} }{1 + {x}^{2} } \bigg)$

Differentiating both sides, w.r.t x, we get,

$\sf \: \frac{dy}{dx} = \frac{1}{ 1 + \bigg( \dfrac{1 - {x}^{2} }{1 + {x}^{2} } \bigg) ^{2} }. \frac{d}{dx} \bigg( \frac{1 - {x}^{2} }{1 + {x}^{2} } \bigg)$

$\sf \: \implies \frac{dy}{dx} = \frac{1}{ 1 + \dfrac{(1 - {x}^{2} )^{2} }{(1 + {x}^{2} )^{2} } }. \bigg \{ \frac{(1 + {x}^{2}) . \frac{d}{dx} (1 - {x}^{2}) - (1 - {x}^{2}). \frac{d}{dx}(1 + {x}^{2} ) }{(1 + {x}^{2} ) ^{2} } \bigg \}$

$\sf \: \implies \frac{dy}{dx} = \frac{(1 + {x}^{2} ) ^{2} }{ (1 + {x}^{2}) + (1 - {x}^{2} )^{2} }. \bigg \{ \frac{(1 + {x}^{2}) . ( - 2x) - (1 - {x}^{2}). ( 2x ) }{(1 + {x}^{2} ) ^{2} } \bigg \}$

$\sf \: \implies \frac{dy}{dx} = \frac{(1 + {x}^{2} ) ^{2} }{ 2 \{(1) ^{2} + ( {x}^{2} )^{2} \} }. \bigg \{ \frac{ - 2x - 2 {x}^{3} - 2x + 2{x}^{3} }{(1 + {x}^{2} ) ^{2} } \bigg \}$

$\sf \: \implies \frac{dy}{dx} = \frac{(1 + {x}^{2} ) ^{2} }{ 2 \{(1) ^{2} + ( {x ^{2} })^{2} \} }. \bigg \{ \frac{ - 2x - 2 {x}^{3} - 2x + 2{x}^{3} }{(1 + {x}^{2} ) ^{2} } \bigg \}$

$\sf \: \implies \frac{dy}{dx} = \frac{(1 + {x}^{2} ) ^{2} }{ 2 (1 + x ^{4} ) }. \bigg \{ \frac{ - 2x - 2x }{(1 + {x}^{2} ) ^{2} } \bigg \}$

$\sf \: \implies \frac{dy}{dx} = \frac{(1 + {x}^{2} ) ^{2} }{ 2 (1 + x ^{4} ) }. \frac{( - 4x ) }{(1 + {x}^{2} ) ^{2} }$

$\sf \: \implies \frac{dy}{dx} = \frac{1 }{ 2 (1 + x ^{4} ) }. \frac{( - 4x ) }{1 }$

$\sf \: \implies \frac{dy}{dx} = \frac{ - 4x}{ 2 (1 + x ^{4} ) }$

$\sf \: \implies \frac{dy}{dx} = \frac{ - 2x}{ 1 + x ^{4} }$