A) \[\frac{1}{x{{y}^{3}}}\]
B) \[\frac{1}{{{x}^{3}}y}\]
C) \[-\frac{1}{x{{y}^{3}}}\]
D) \[-\frac{1}{{{x}^{3}}y}\]
Correct Answer: B
Solution :
Given, \[{{x}^{2}}+{{y}^{2}}=t-\frac{1}{t}\] and \[{{x}^{4}}+{{y}^{4}}={{t}^{2}}+\frac{1}{{{t}^{2}}}\] \[\Rightarrow \] \[{{x}^{4}}+{{y}^{4}}={{\left( t-\frac{1}{t} \right)}^{2}}+2\] \[={{({{x}^{2}}+{{y}^{2}})}^{2}}+2\] \[\Rightarrow \] \[{{x}^{4}}+{{y}^{4}}={{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}+2\] \[\Rightarrow \] \[2{{x}^{2}}{{y}^{2}}=-2\] \[\Rightarrow \] \[{{x}^{2}}{{y}^{2}}=-1\] \[\Rightarrow \] \[{{y}^{2}}=-\frac{1}{{{x}^{2}}}\] On differentiating w.r.t.\[x,\]we get \[2y\frac{dy}{dx}=\frac{2}{{{x}^{3}}}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1}{{{x}^{3}}y}\]
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