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प्रश्न
If \[x^{16} y^9 = \left( x^2 + y \right)^{17}\] ,prove that \[x\frac{dy}{dx} = 2 y\] ?
उत्तर
\[\text{ We have}, x^{16} y^9 = \left( x^2 + y \right)^{17} \]
Taking log on both sides,
\[\log\left( x^{16} y^9 \right) = \log \left( x^2 + y \right)^{17} \]
\[ \Rightarrow 16\log x + 9\log y = 17\log\left( x^2 + y \right)\]
Differentiating with respect to x using chain rule,
\[16\frac{d}{dx}\left( \log x \right) + 9\frac{d}{dx}\left( \log y \right) = 17\frac{d}{dx}\log\left( x^2 + y \right)\]
\[ \Rightarrow \frac{16}{x} + \frac{9}{y}\frac{dy}{dx} = \frac{17}{x^2 + y}\frac{d}{dx}\left( x^2 + y \right)\]
\[ \Rightarrow \frac{16}{x} + \frac{9}{y}\frac{dy}{dx} = \frac{17}{x^2 + y}\left[ 2x + \frac{dy}{dx} \right]\]
\[ \Rightarrow \frac{9}{y}\frac{dy}{dx} - \frac{17}{x^2 + y}\frac{dy}{dx} = \frac{34x}{x^2 + y} - \frac{16}{x}\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{9}{y} - \frac{17}{x^2 + y} \right] = \frac{34x}{x^2 + y} - \frac{16}{x}\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{9\left( x^2 + y \right) - 17y}{y\left( x^2 + y \right)} \right] = \left[ \frac{34 x^2 - 16\left( x^2 + y \right)}{x\left( x^2 + y \right)} \right]\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{9 x^2 + 9y - 17y}{y\left( x^2 + y \right)} \right] = \left[ \frac{34 x^2 - 16 x^2 - 16y}{x\left( x^2 + y \right)} \right]\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{9 x^2 - 8y}{y\left( x^2 + y \right)} \right] = \left[ \frac{18 x^2 - 16y}{x\left( x^2 + y \right)} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y}{x}\left[ \frac{2\left( 9 x^2 - 8y \right)}{9 x^2 - 8y} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2y}{x}\]
\[ \Rightarrow x\frac{dy}{dx} = 2y\]
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