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If Y = [ Log ( X + √ X 2 + 1 ) ] 2 Show that ( 1 + X 2 ) D 2 Y D X 2 + X D Y D X = 2 ? - Mathematics

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प्रश्न

If \[y = \left[ \log \left( x + \sqrt{x^2 + 1} \right) \right]^2\] show that \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 2\] ?

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उत्तर

Here,

\[y = \left[ \log\left( x + \sqrt{x^2 + 1} \right) \right]^2 \]

\[\text { Differentiating w . r . t . x, we get }\]

\[\frac{d y}{d x} = \frac{2\log\left( x + \sqrt{x^2 + 1} \right)}{\left( x + \sqrt{x^2 + 1} \right)} \times \left( 1 + \frac{2x}{2\sqrt{x^2 + 1}} \right)\]

\[ \Rightarrow \frac{d y}{d x} = \frac{2\log\left( x + \sqrt{x^2 + 1} \right)}{\left( x + \sqrt{x^2 + 1} \right)} \times \left( \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}} \right)\]

\[ \Rightarrow \frac{d y}{d x} = \frac{2\log\left( x + \sqrt{x^2 + 1} \right)}{\sqrt{x^2 + 1}}\]

\[\text { Differentiating again w . r . t . x, we get }\]

\[\frac{d^2 y}{d x^2} = \frac{2 - \frac{2x\log\left( x + \sqrt{x^2 + 1} \right)}{\sqrt{x^2 + 1}}}{x^2 + 1}\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{2 - x\frac{dy}{dx}}{x^2 + 1}\]

\[ \Rightarrow \left( x^2 + 1 \right)\frac{d^2 y}{d x^2} = 2 - x\frac{dy}{dx}\]

\[ \Rightarrow \left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 2\]

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पाठ 12: Higher Order Derivatives - Exercise 12.1 [पृष्ठ १७]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 12 Higher Order Derivatives
Exercise 12.1 | Q 27 | पृष्ठ १७

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