If \[x=\frac{2\,t}{1+{{t}^{2}}},\,\,y=\frac{1-{{t}^{2}}}{1+{{t}^{2}}},\]then \[\frac{d\,y}{d\,x}\] equals [RPET 1999]

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

1.	If \[x=\frac{2\,t}{1+{{t}^{2}}},\,\,y=\frac{1-{{t}^{2}}}{1+{{t}^{2}}},\]then \[\frac{d\,y}{d\,x}\] equals [RPET 1999]
A.	\[\frac{2\,t}{{{t}^{2}}+1}\]
B.	\[\frac{2\,t}{{{t}^{2}}-1}\]
C.	\[\frac{2\,t}{1-{{t}^{2}}}\]
D.	None of these
Answer» C. \[\frac{2\,t}{1-{{t}^{2}}}\]

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If \[x=\frac{2\,t}{1+{{t}^{2}}},\,\,y=\frac{1-{{t}^{2}}}{1+{{t}^{2}}},\]then \[\frac{d\,y}{d\,x}\] equals [RPET 1999]

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