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If Log √ X 2 + Y 2 = Tan − 1 ( Y X ) Prove that D Y D X = X + Y X − Y ? - Mathematics

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प्रश्न

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

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उत्तर

\[\text{ We have }, \log\sqrt{x^2 + y^2} = \tan^{- 1} \left( \frac{x}{y} \right)\]

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

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

Differentiate with respect to x, we get,

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

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

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

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

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

\[ \Rightarrow x + y\frac{d y}{d x} = x\frac{d y}{d x} - y\]

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

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

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

\[ \Rightarrow \frac{d y}{d x} = \frac{x + y}{x - y}\]

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पाठ 11: Differentiation - Exercise 11.04 [पृष्ठ ७५]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.04 | Q 17 | पृष्ठ ७५

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