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प्रश्न
If \[y^x + x^y + x^x = a^b\] ,find \[\frac{dy}{dx}\] ?
उत्तर
\[\text{ Given that }y^x + x^y + x^x = a^b \]
\[\text{ Putting u }= y^x , v = x^y \text{and }w = x^x , \text{ we get }\]
\[ u + v + w = a^b \]
\[ \therefore \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0 . . . \left( i \right)\]
\[\text{ Now, u } = y^x \]
Taking log on both sides,
\[\log u = x \log y\]
\[\Rightarrow \frac{1}{u}\frac{du}{dx} = x\frac{d}{dx}\left( \log y \right) + \log y\frac{d}{dx}\left( x \right) \left[ \text{ using product } rule \right]\]
\[ \Rightarrow \frac{1}{u}\frac{du}{dx} = x\frac{1}{y}\frac{dy}{dx} + \log y \times 1\]
\[ \Rightarrow \frac{du}{dx} = u\left( \frac{x}{y}\frac{dy}{dx} + \log y \right)\]
\[ \Rightarrow \frac{du}{dx} = y^x \left( \frac{x}{y}\frac{dy}{dx} + \log y \right) . . . \left( ii \right)\]
\[\text{ Also, v } = x^y\]
Taking log on both sides,
\[\log v = y \log x\]
\[\Rightarrow \frac{1}{v}\frac{dv}{dx} = y\frac{d}{dx}\left( \log x \right) + \log x\frac{dy}{dx}\]
\[ \Rightarrow \frac{1}{v}\frac{dv}{dx} = y\frac{1}{x} + \log x\frac{dy}{dx}\]
\[ \Rightarrow \frac{dv}{dx} = v\left[ \frac{y}{x} + \log x\frac{dy}{dx} \right]\]
\[ \Rightarrow \frac{dv}{dx} = x^y \left[ \frac{y}{x} + \log x\frac{dy}{dx} \right] . . . \left( iii \right)\]
\[\text{ Again, w } = x^x\]
Taking log on both sides,
\[\log w = x \log x\]
\[\Rightarrow \frac{1}{w}\frac{dw}{dx} = x\frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( x \right)\]
\[ \Rightarrow \frac{1}{w}\frac{dw}{dx} = x\frac{1}{x} + \log x\left( 1 \right)\]
\[ \Rightarrow \frac{dw}{dx} = w\left( 1 + \log x \right)\]
\[ \Rightarrow \frac{dw}{dx} = x^x \left( 1 + \log x \right) . . . \left( iv \right)\]
\[\text{ From } \left( i \right), \left( ii \right), \left( iii \right)\text{ and }\left( iv \right), \text{ we have }\]
\[ y^x \left( \frac{x}{y}\frac{dy}{dx} + \log y \right) + x^y \left( \frac{y}{x} + \log x\frac{dy}{dx} \right) + x^x \left( 1 + \log x \right) = 0\]
\[ \Rightarrow \left( x . y^{x - 1} + x^y . \log x \right)\frac{dy}{dx} = - x^x \left( 1 + \log x \right) - y . x^{y - 1} - y^x \log y\]
\[ \therefore \frac{dy}{dx} = \frac{- \left\{ y^x \log y + y . x^{y - 1} + x^x \left( 1 + \log x \right) \right\}}{x . y^{x - 1} + x^y \log x}\]
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