Question

D/dx log (x+√x^2+a^2)-log (-x+√x^2+a^2) = ?

Answer 1

Answer:

$\dfrac{2}{ \sqrt{ \: ( {x}^{2} \: + {a}^{2} \: ) } }$

Step-by-step explanation:

To find ---->

$derivative \: of \: \\ log \: ( \: x \: + \: \sqrt{ {x}^{2} \: + \: {a}^{2} \: } \: ) \: - \: log \: ( - x \: + \sqrt{ {x}^{2} \: + \: {a}^{2} } \: )$

Concept used ---->

1)

$\dfrac{d}{dx} \: ( \: logx \: ) = \dfrac{1}{x}$

2)

$\dfrac{d}{dx} \: ( \: \sqrt{x} \: ) = \: \dfrac{1}{2 \: \sqrt{x} }$

3)

$\dfrac{ {d} }{dx} \: ( \: {x}^{2} \: ) \: = 2x$

4)

$\dfrac{d}{dx} \: ( \: constant \: ) = \: 0$

Solution----> Let,

$y \: = log(x + \sqrt{ {x}^{2} + {a}^{2} } \: ) - log( - x + \sqrt{ {x}^{2} + {a}^{2} } \: )$

$differentiating \: with \: respect \: to \: x \: we \: get$

$\dfrac{dy}{dx} = \dfrac{d}{dx}log(x + \sqrt{ {x}^{2} + {a}^{2} } \: ) - \dfrac{d}{dx}log( - x + \sqrt{ {x}^{2} + {a}^{2} } \: )$

$= \dfrac{ \: \dfrac{d}{dx}(x + \sqrt{ {x}^{2} + {a}^{2} } \: ) }{x + \sqrt{ {x}^{2} + {a}^{2} } } - \dfrac{ \dfrac{d}{dx}( - x + \sqrt{ {x}^{2} + {a}^{2} } \: ) }{ - x + \sqrt{ {x}^{2} + {a}^{2} } }$

$= \dfrac{1 + \dfrac{1}{2 \sqrt{ {x}^{2} + {a}^{2} } } \: \dfrac{d}{dx} \: ( {x}^{2} + {a \: }^{2} ) }{x + \sqrt{ {x}^{2} + {a}^{2} } } - \dfrac{ - 1 + \dfrac{1}{2 \sqrt{ {x}^{2} + {a}^{2} } } \dfrac{d}{dx}( {x}^{2} + {a}^{2}) }{( - x + \sqrt{ {x}^{2} + {a}^{2} }) }$

$= \dfrac{1 + \dfrac{2x}{2 \sqrt{ {x}^{2} + {a}^{2} } } }{x + \sqrt{ {x}^{2} + {a}^{2} } } - \dfrac{ - 1 + \dfrac{2x}{2 \sqrt{ {x}^{2} + {a}^{2} } } }{ - x + \sqrt{ {x}^{2} + {a}^{2} } }$

$= \dfrac{ \dfrac{ \sqrt{ {x}^{2} + {a}^{2} } + x }{ \sqrt{ {x}^{2} + {a}^{2} } } }{x + \sqrt{ {x}^{2} + {a}^{2} } } - \dfrac{ \dfrac{ - \sqrt{ {x}^{2} + {a}^{2} } + x }{ \sqrt{ {x}^{2} + {a}^{2} } } }{ - (x - \sqrt{ {x}^{2} + {a}^{2} }) }$

$= \dfrac{1}{ \sqrt{ {x}^{2} + {a}^{2} } } + \dfrac{1}{ \sqrt{ {x}^{2} + {a}^{2} } }$

$= \dfrac{2}{ \sqrt{ {x}^{2} + {a}^{2} } }$

Answer 2

$\huge\boxed{\fcolorbox{violet}{violet}{Answer}}$