Differentiation Question 440
Question: If $ y^{x}+x^{y}=a^{b} $ ,then $ \frac{dy}{dx}= $
Options:
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}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ x^{y}+y^{x}=a^{b} $ ; Let $ x^{y}=u $ and $ y^{x}=v $
Therefore $ u+v=a^{b} $
Therefore $ \frac{du}{dx}+\frac{dv}{dx}=0 $
Now differentiating both by taking their $ \log $ separately $ \frac{du}{dx}=x^{y}( \frac{y}{x}+\frac{dy}{dx}\log x ) $
……..(i) and $ \frac{dv}{dx}=y^{x}( \log y+\frac{x}{y}.\frac{dy}{dx} ) $
……..(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}}} $ .
BETA
