A) \[\frac{{{x}^{2}}}{{{y}^{2}}}\]
B) \[{{x}^{2}}{{y}^{2}}\]
C) \[\frac{{{y}^{2}}}{{{x}^{2}}}\]
D) \[\frac{-{{y}^{2}}}{{{x}^{2}}}\]
Correct Answer: D
Solution :
Given that, \[y=1+\frac{1}{x}+\frac{1}{{{x}^{2}}}+\frac{1}{{{x}^{3}}}+....\] \[\therefore \] \[y=\frac{1}{1-\frac{1}{x}}\] (GP series) \[=\frac{x}{x-1}\] ...(i) On differentiating w.r.t.\[x,\]we get \[\frac{dy}{dx}=\frac{1.(x-1)-x.1}{{{(x-1)}^{2}}}=-\frac{1}{{{(x-1)}^{2}}}\] \[\Rightarrow \] \[\frac{dy}{dx}=-\frac{{{y}^{2}}}{{{x}^{2}}}\] [from(i)]You need to login to perform this action.
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