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

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

1.	If \[x=\frac{1-{{t}^{2}}}{1+{{t}^{2}}}\]and \[y=\frac{2at}{1+{{t}^{2}}}\], then \[\frac{dy}{dx}=\]
A.	\[\frac{a(1-{{t}^{2}})}{2t}\]
B.	\[\frac{a({{t}^{2}}-1)}{2t}\]
C.	\[\frac{a({{t}^{2}}+1)}{2t}\]
D.	\[\frac{a({{t}^{2}}-1)}{t}\]
Answer» C. \[\frac{a({{t}^{2}}+1)}{2t}\]

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

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