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If Xy = 4, Prove that X ( D Y D X + Y 2 ) = 3 Y ? - Mathematics

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प्रश्न

If xy = 4, prove that \[x\left( \frac{dy}{dx} + y^2 \right) = 3 y\] ?

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उत्तर

\[\text{We have, xy } = 4\]

\[ \Rightarrow y = \frac{4}{x}\]

Differentiate it with respect to x,

\[\frac{d y}{d x} = \frac{d}{dx}\left( \frac{4}{x} \right)\]

\[ \Rightarrow \frac{d y}{d x} = 4\frac{d}{dx}\left( x^{- 1} \right)\]

\[ \Rightarrow \frac{d y}{d x} = 4\left( - 1 \times x^{- 1 - 1} \right)\]

\[ \Rightarrow \frac{d y}{d x} = 4\left( - \frac{1}{x^2} \right)\]

\[ \Rightarrow \frac{d y}{d x} = \frac{- 4}{x^2}\]

\[ \Rightarrow \frac{d y}{d x} = - \frac{4}{\left( \frac{4}{y} \right)^2} \left[ \because x = \frac{4}{y} \right]\]

\[ \Rightarrow \frac{d y}{d x} = - \frac{4 y^2}{16}\]

\[ \Rightarrow \frac{d y}{d x} = - \frac{y^2}{4}\]

\[ \Rightarrow 4\frac{d y}{d x} = - y^2 \]

\[ \Rightarrow 4\frac{d y}{d x} = 3 y^2 - 4 y^2 \]

\[ \Rightarrow 4\frac{d y}{d x} + 4 y^2 = 3 y^2 \]

\[ \Rightarrow 4\left( \frac{d y}{d x} + y^2 \right) = 3 y^2 \]

Dividing both side by x,

\[\Rightarrow \frac{4}{x}\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{x}\]

\[ \Rightarrow y\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{x} \]

\[ \Rightarrow x\left( \frac{d y}{d x} + y^2 \right) = \frac{3 y^2}{y}\]

\[ \Rightarrow x\left( \frac{d y}{d x} + y^2 \right) = 3y\]

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पाठ 11: Differentiation - Exercise 11.02 [पृष्ठ ३८]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.02 | Q 73 | पृष्ठ ३८

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