" Let a function "y" -y "(x)" be defined parametrically by "x=2t*||y|=-z^(2)+y|y" .Then "y'(x)x=0

Let a function y=y(x) be defined parametrically by x=2t-|t|y=t^(2)+t|t| then y'(x),x>0

Let a function y = y (x) be defined parametrically by x = 2t - |t|, y =t^2 +t|t| . Then y^(1) (x), x gt 0

If y = f(x) defined parametrically by x = 2t - |t - 1| and y = 2t^(2) + t|t| , then

Let y = f(x) be defined parametrically as y = t^(2) + t|t|, x = 2t - |t|, t in R . Then, at x = 0,find f(x) and discuss continuity.

Let y = f(x) be defined parametrically as y = t^(2) + t|t|, x = 2t - |t|, t in R . Then, at x = 0,find f(x) and discuss continuity.

Consider a function y = f(x) defined parametrically as y=2t+t, x=t^2 foralltin R then the function f(x) is