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