Question

If $\sf y = (x+ \sqrt{1+x^{2}})^{n}$, then $\sf (1+x)^{2} \dfrac{d^{2}y}{dx^{2}}+ x\dfrac{dy}{dx}$ is?

Answer 1

Answer:

I posted the answer in attachment.

Hope it helps u mam /sir .

I tried my best .

Thank u.

Answer 2

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▪ Given :-

$\mathtt{y = (x+ \sqrt{1+x^{2}})^{n}}$

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To Find :-

$\mathtt{(1+x)^{2} \dfrac{d^{2}y}{dx^{2}}+ x\dfrac{dy}{dx}}$

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Answer :-

$\huge\purple{ \bf n {}^{2} y}$

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Step by step Solution :-

$\mathtt{ y = (x+ \sqrt{1+x^{2}})^{n}}$

Differentiating both sides w.r.t x

$\large\mathtt{\dfrac{dy}{dx} = n.(x+ \sqrt{1+x^{2}})^{n - 1} } \\ \large\mathtt{ \times \frac{d}{dx} (x+ \sqrt{1+x^{2}})} \\ \\ \large \mathtt{\dfrac{dy}{dx} = n.(x+ \sqrt{1+x^{2}})^{n - 1} } \\ \large \mathtt{ \times \frac{d}{dx} (x+ {(1+x^{2} {)}^{1/2} }) {}^{} } \\ \\ \large \mathtt{\dfrac{dy}{dx} = n.(x+ \sqrt{1+x^{2}})^{n - 1} } \\ \large\mathtt{ \times (1 + \frac{d}{dx}(1+x^{2} {)}^{1/2})} \\ \\ \large\mathtt{\dfrac{dy}{dx} = n.(x+ \sqrt{1+x^{2}})^{n - 1} } \\ \large \mathtt{ \times (1 + \frac{1}{2} (1+x^{2} {)}^{ - 1/2}\times 2x})$

$\large\mathtt{\dfrac{dy}{dx} = n.(x+ \sqrt{1+x^{2}})^{n - 1} } \\ \large \mathtt{ \times \bigg(1 + \frac{x}{ \sqrt{1 + {x}^{2} } } } \bigg) \\ \\ \large \mathtt{\dfrac{dy}{dx} = n.(x+ \sqrt{1+x^{2}})^{n - 1} } \\ \large\mathtt{ \times \bigg( \frac{ \sqrt{1 + {x}^{2} } + x }{ \sqrt{1 + {x}^{2} } } } \bigg) \\ \\ \large= \mathtt{ \dfrac{n.(x + \sqrt{1 + {x}^{2} } ) {}^{n} }{ \sqrt{ 1 + {x}^{2} } } } \\ \\ \large \mathtt{ \frac{dy}{dx} = \frac{ny}{ \sqrt{1 + {x}^{2} } } }$

$\purple{: \longmapsto\boxed{\boxed{ \large\mathtt{ ( \sqrt{1 + {x}^{2} } ) \dfrac{dy}{dx} = ny}}}}\:\:---(1)$

Differentiating both sides w.r.t x

$\large\mathtt{({\sqrt{ 1 + {x}^{2} }). \dfrac{ {{d}^{2}y }}{ {{dx}^{2} }}}} \\ \\ \large + \mathtt{ \frac{dy}{dx}. \frac{d}{dx}( \sqrt{1 + {x}^{2} } ) = n \frac{dy}{dx} }$

$\mathtt{ \sqrt{1 + {x}^{2} }. \dfrac{ {d}^{2} y}{dx {}^{2} } \: \: } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: + \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = n \frac{dy}{dx}} \\ \mathtt{ \dfrac{dy}{dx} . \frac{1}{ \not2} \times \frac{ \not2x}{ \sqrt{1 + {x}^{2} } } }$

$\large\bf Multiplying \:\: Both \:\: Sides \:\: by\: \\ \large \bf \sqrt{1 + {x}^{2} }$

We Get ,

$\large\mathtt{(1 + {x}^{2}) \dfrac{ {d}^{2} y}{dx {}^{2} } + x \dfrac{dy}{dx} = n.(ny)} \\ \bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \{\because {eq}^{n} \: \: (1) \}$

$\large\pink{ : \longmapsto\mathtt{(1 + {x}^{2}) \dfrac{ {d}^{2} y}{dx {}^{2} } + x \dfrac{dy}{dx} = n {}^{2} y}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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