Question

Find the derivative :
$\sf y= \{ x + \sqrt{ {x}^{2} + {a}^{2} } \: \}^{n} \\ \\ \bf find \: \: \frac{dy}{dx}$
​

Answer 1

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

${\implies{\sf{\green{y = \big( x + \sqrt{ {x}^{2} + {a}^{2} }\big) ^{n} }}}}$

${\bf{\blue{\underline{Now:}}}}$

${\sf{\underline{\blue{ \star \: differentiate \: w.r.t.x,}}}}$

${\bf{ \frac{dy}{dx} = n \big( x + \sqrt{ {x}^{2} + {a}^{2} } \big) ^{n - 1} \frac{d}{dx} \big(x + \sqrt{ {x}^{2} + {a}^{2} } \big)}}$

${\implies{\bf{ n \big(x + \sqrt{ {x}^{2} + {a}^{2} } \big) ^{n - 1} \times \big [ 1 + ( {x}^{2} + {a}^{2}) ^{ \frac{1}{2} } \big] }}}$

${\implies{\bf{ n \big(x + \sqrt{ {x}^{2} + {a}^{2} } \big) ^{n - 1} \times \big [ 1 + \frac{1}{2 \sqrt{ {x}^{2} + {a}^{2} } } \ \times \frac{d}{dx} ( {x}^{2} + {a}^{2}) \big] }}}$

${\implies{\bf{ n \big(x + \sqrt{ {x}^{2} + {a}^{2} } \big) ^{n - 1} \times \big [ 1 + \frac{1}{2 \sqrt{ {x}^{2} + {a}^{2} } } \ \times (2x + 0) \big] }}}$

${\implies{\bf{ n \big(x + \sqrt{ {x}^{2} + {a}^{2} } \big) ^{n - 1} \times \big [ \frac{ \sqrt{ {x}^{2} + {a}^{2} } + x}{ \sqrt{ {x}^{2} + {a}^{2} } } \big] }}}$

${\implies{\bf{ \frac{n \big(x + \sqrt{ {x}^{2} + {a}^{2} } \big) ^{n} }{ \sqrt{ {x}^{2} + {a}^{2} } } }}}$

${\bf{\blue{hence \: \frac{dy}{dx} = \frac{n(x + \sqrt{ {x}^{2} + {a}^{2} } )^{n} }{ \sqrt{ {x}^{2} + {a}^{2} } } }}}$

${\orange{ \boxed{{ \sf{ \star \: Formula \: used}}}}}$

${\green{{ \sf{ \star \: \frac{d}{dx}( {x}^{n) } = n {x}^{n - 1} }}}}$

Answer 2

Step-by-step explanation:

$dy \div dx = n(x + \sqrt{x { + a {) {n \div \sqrt{x { + a {?}^{2} }^{2} } }^{?} }^{2} }^{2} }$

have a great day NANBA