A) \[\frac{xy}{(1-x)}\]
B) \[\frac{(1+y)}{(1-x)}\]
C) \[\frac{y}{(1-xy)}\]
D) \[\frac{-1}{{{(x-1)}^{2}}}\]
E) \[\frac{1}{({{x}^{2}}-1)}\]
Correct Answer: D
Solution :
\[xy=x+y\Rightarrow xy-y=x\] \[\Rightarrow \] \[y(x-1)=x\] \[\Rightarrow \] \[yx=x+y\] On differentiating w.r.t.\[x,\]we get \[x\frac{dy}{dx}+y=1+\frac{dy}{dx}\] \[\Rightarrow \] \[\frac{dy}{dx}(x-1)=1-y\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1-y}{x-1}=\frac{1-\frac{x}{x-1}}{x-1}\] \[=\frac{x-1-x}{{{(x-1)}^{2}}}=\frac{-1}{{{(x-1)}^{2}}}\]
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