If \[{{x}^{2}}+{{y}^{2}}=t-\frac{1}{t},\]\[{{x}^{4}}+{{y}^{4}}={{t}^{2}}+\frac{1}{{{t}^{2}}}\], then \[{{x}^{3}}y\frac{dy}{dx}=\]

If \[{{x}^{2}}+{{y}^{2}}=t-\frac{1}{t},\]\[{{x}^{4}}+{{y}^{4}}={{t}^{2

1.	If \[{{x}^{2}}+{{y}^{2}}=t-\frac{1}{t},\]\[{{x}^{4}}+{{y}^{4}}={{t}^{2}}+\frac{1}{{{t}^{2}}}\], then \[{{x}^{3}}y\frac{dy}{dx}=\]
A.	1
B.	2
C.	3
D.	4
Answer» B. 2

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If \[{{x}^{2}}+{{y}^{2}}=t-\frac{1}{t},\]\[{{x}^{4}}+{{y}^{4}}={{t}^{2}}+\frac{1}{{{t}^{2}}}\], then \[{{x}^{3}}y\frac{dy}{dx}=\]

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