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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

EXPLANATION.

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

As we know that,

Formula of :

$\sf \implies \dfrac{d}{dx} [f(x).g(x)] = f(x). \dfrac{d}{dx} [g(x)] + g(x). \dfrac{d}{dx} [f(x)].$

Using this formula in the equation, we get.

$\sf \implies \dfrac{dy}{dx} = \bigg[ (x) .\dfrac{d}{dx} \sqrt{a^{2}+ x^{2} } \ + \sqrt{a^{2}+ x^{2} } . \dfrac{d}{dx} (x) \bigg] + \bigg[ \dfrac{a^{2} }{x + \sqrt{a^{2} + x^{2} } } . \dfrac{d}{dx} ( x + \sqrt{a^{2}+ x^{2} ) } \bigg]$

$\sf \implies \dfrac{dy}{dx} = \bigg[ x . \dfrac{2x}{2\sqrt{a^{2}+ x^{2} } } +\sqrt{a^{2} + x^{2} } } \bigg] + \bigg[ \dfrac{a^{2} }{x + \sqrt{a^{2} + x^{2} } } \bigg] \bigg[ 1 + \dfrac{2x}{2\sqrt{a^{2}+ x^{2} } } \bigg]$

$\sf \implies \dfrac{dy}{dx} = \bigg[ \dfrac{x^{2} }{\sqrt{a^{2} + x^{2} } } + \sqrt{a^{2}+ x^{2} } \bigg] + \bigg[ \dfrac{a^{2} }{x + \sqrt{a^{2}+ x^{2} } } \bigg] \bigg[ \dfrac{\sqrt{a^{2} + x^{2} } + x}{\sqrt{a^{2} + x^{2} } } \bigg]$

$\sf \implies \dfrac{dy}{dx} = \bigg[ \dfrac{x^{2} }{\sqrt{a^{2} + x^{2} } } + \sqrt{a^{2} + x^{2} } \ + \dfrac{a^{2} }{\sqrt{a^{2}+ x^{2} } } \bigg]$

$\sf \implies \dfrac{dy}{dx} = \bigg[ \dfrac{a^{2} + x^{2} }{\sqrt{a^{2}+ x^{2} } } + \sqrt{a^{2} + x^{2} } \bigg]$

$\sf \implies \dfrac{dy}{dx} = \sqrt{a^{2}+ x^{2} } + \sqrt{a^{2}+ x^{2} }$

$\sf \implies \dfrac{dy}{dx} = 2\sqrt{a^{2} + x^{2} }$

Hence Proved.