A) -1
B) 0
C) 1
D) None of these
Correct Answer: A
Solution :
\[{{x}^{y}}\cdot {{y}^{x}}=16\] \[\therefore \] \[{{\log }_{e}}{{x}^{y}}+{{\log }^{e}}{{y}^{x}}={{\log }_{e}}\,16\] \[\Rightarrow \] \[y\,{{\log }_{e}}x+x{{\log }_{e}}y=4{{\log }_{e}}2\] Now, on differentiating both sides, w.r.t. x, we get \[\frac{y}{x}+{{\log }_{e}}x\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}+{{\log }_{e}}y\cdot 1=0\] \[\therefore \] \[\frac{dy}{dx}=-\frac{\left( {{\log }_{e}}y+\frac{y}{x} \right)}{\left( {{\log }_{e}}x+\frac{x}{y} \right)}\] \[\therefore \] \[{{\left( \frac{dy}{dx} \right)}_{(2,\,2)}}=-\,\frac{({{\log }_{e}}2+1)}{({{\log }_{e}}2+1)}=-1\]
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