A) \[\frac{1}{{{x}^{2}}{{y}^{3}}}\]
B) \[\frac{1}{x{{y}^{3}}}\]
C) \[\frac{1}{{{x}^{2}}{{y}^{2}}}\]
D) \[\frac{1}{{{x}^{3}}y}\]
E) \[\frac{-1}{{{x}^{3}}y}\]
Correct Answer: D
Solution :
Given, \[{{x}^{2}}+{{y}^{2}}=t-\frac{1}{t}\]and \[{{x}^{4}}+{{y}^{4}}={{t}^{2}}+\frac{1}{{{t}^{2}}}\] \[\Rightarrow \] \[{{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}={{t}^{2}}+\frac{1}{{{t}^{2}}}-2\] \[\Rightarrow \] \[{{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}={{x}^{2}}+{{y}^{4}}-2\] \[\Rightarrow \] \[{{x}^{2}}{{y}^{2}}+1=0\] \[\Rightarrow \] \[{{y}^{2}}=\frac{-1}{{{x}^{2}}}\] On differentiating w.r.t.\[x,\]we get \[2y\frac{dy}{dx}=\frac{2}{{{x}^{3}}}\Rightarrow \frac{dy}{dx}=\frac{1}{{{x}^{3}}y}\]
You need to login to perform this action.
You will be redirected in
3 sec
