If \[x=\frac{1-{{t}^{2}}}{1+{{t}^{2}}}\]and \[y=\frac{2t}{1+{{t}^{2}}}\], then \[\frac{dy}{dx}=\] [Karnataka CET 2000; Pb. CET 2002]

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

1.	If \[x=\frac{1-{{t}^{2}}}{1+{{t}^{2}}}\]and \[y=\frac{2t}{1+{{t}^{2}}}\], then \[\frac{dy}{dx}=\] [Karnataka CET 2000; Pb. CET 2002]
A.	\[\frac{-y}{x}\]
B.	\[\frac{y}{x}\]
C.	\[\frac{-x}{y}\]
D.	\[\frac{x}{y}\]
Answer» D. \[\frac{x}{y}\]

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If \[x=\frac{1-{{t}^{2}}}{1+{{t}^{2}}}\]and \[y=\frac{2t}{1+{{t}^{2}}}\], then \[\frac{dy}{dx}=\] [Karnataka CET 2000; Pb. CET 2002]

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