The derivative of the function represented parametrically as `x=2t=|t|,y=t^3+t^2|t|a tt=0` is a. -1 b. 1 c. 0 d. does not exist

Options

A. `-1`
B. 0
C. 1
D. does not exist

Answer

B

Explanation

We have,
`x=2t-|t|, y=t^(3)+t^(2)|t|`
`rArr" "x=3t, y=0` when t lt 0
X = t, `y=2t^(3)` when `x ge 0`
Eliminating the parameter t, we get
`y={{:(0",",xlt0),(2x^(3)",",xge0):}` ltbr. Differentiating w.r.t. x, we get
`(dy)/(dx)={{:(0",",xlt0),(6x^(2)",",xge0):}`
Hence, the function is differentiable every where and its derivative at x = 0 (t = 0) is 0.