Question

differentiation of tan^-1{x/✓(1+x²)}​

Answer 1

Step-by-step explanation:

${ \frac{d}{dx} (\tan^{ - 1} ( \frac{x}{ \sqrt{1 + {x}^{2} } } ))} \\ = (\frac{1}{1 + (\frac{x}{ \sqrt{1 + {x}^{2} } } ) ^{2} } ) \times \frac{1 \sqrt{1 + {x}^{2} } - \frac{x}{ 2\sqrt{1 + {x}^{2} } }(2x) }{1 + {x}^{2} } \\ = \frac{1}{1 + ( \frac{ {x}^{2} }{1 + {x}^{2} } )} \times ( \frac{ \sqrt{1 + {x}^{2} } - \frac{ {x}^{2} }{ \sqrt{1 + {x}^{2} } } }{1 + {x}^{2} } ) \\ = (\frac{1 + {x}^{2} }{1 + 2 {x}^{2} } ) \times ( \frac{ \frac{1 + {x }^{2} - {x}^{2} }{ \sqrt{1 + {x}^{2} } }}{1 + {x}^{2} } ) \\ = ( \frac{1 + {x}^{2} }{1 + 2 {x}^{2} } ) \times ( \frac{1}{(1 + {x}^{2}) ( \sqrt{1 + {x}^{2} } )}) \\ \\ = \frac{1}{(1 + 2 {x}^{2})( \sqrt{1 + {x}^{2} }) } \\ \\ ans$

Hlo mate hope my efforts will help u

Answer 2

Answer:

Sorry But I don't know the Answer