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

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

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.

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