A) \[-\frac{y{{x}^{y-1}}+{{y}^{x}}\log y}{x{{y}^{x-1}}+{{x}^{y}}\log x}\]
B) \[\frac{y{{x}^{y-1}}+{{y}^{x}}\log y}{x{{y}^{x-1}}+{{x}^{y}}\log x}\]
C) \[-\frac{y{{x}^{y-1}}+{{y}^{x}}}{x{{y}^{x-1}}+{{x}^{y}}l}\]
D) \[\frac{y{{x}^{y-1}}+{{y}^{x}}}{x{{y}^{x-1}}+{{x}^{y}}}\]
Correct Answer: A
Solution :
\[{{x}^{y}}+{{y}^{x}}={{a}^{b}}\]; Let \[{{x}^{y}}=u\] and \[{{y}^{x}}=v\] Þ \[u+v={{a}^{b}}\]Þ\[\frac{du}{dx}+\frac{dv}{dx}=0\] Now differentiating both by taking their \[\log \]separately \[\frac{du}{dx}={{x}^{y}}\left( \frac{y}{x}+\frac{dy}{dx}\log x \right)\] ?..(i) and \[\frac{dv}{dx}={{y}^{x}}\left( \log y+\frac{x}{y}.\frac{dy}{dx} \right)\] ?..(ii) Therefore, by (i) and (ii), \[\frac{dy}{dx}=-\frac{y{{x}^{y-1}}+{{y}^{x}}\log y}{{{x}^{y}}\log x+x{{y}^{x-1}}}\].You need to login to perform this action.
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