Question

differentiate e power log (x+ root x square+a square​

Answer 1

Answer:

Question

$\bf \:Differnatiate \: {e}^{ log(x + \sqrt{ {x}^{2} + {a}^{2} } ) } w.r.t. \: x$

☆Answer☆

Formula Used:-

$\bf \:\dfrac{d}{dx} {x}^{n } = n {x}^{n - 1}$

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

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

$\bf \:\dfrac{d}{dx}k = 0$

$\bf \: {e}^{ log(x) } = x$

Solution:-

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

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

$\bf \:Differnatiate \: w.r.t. \: x$

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

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

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

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} + {a}^{2} } } (1 + \dfrac{1}{2 \sqrt{ {x}^{2} + {a}^{2} } }(2x))$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} + {a}^{2} } } (1 + \dfrac{x}{ \sqrt{ {x}^{2} + {a}^{2} } })$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{1}{x + \sqrt{ {x}^{2} + {a}^{2} } } ( \dfrac{x + \sqrt{ {x}^{2} + {a}^{2} } }{ \sqrt{ {x}^{2} + {a}^{2} } })$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{1}{ \sqrt{ {x}^{2} + {a}^{2} } }$

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Answer 2

Step-by-step explanation:

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