If `x^3-2x^2y^2+5x+y-5=0` and `y(1)=1`, then (a)`y^(prime)(1)=4/3` (b) `y'(1)=-4/3` (c)`y''(1)=-8(22)/(27)` (d) `y^(prime)(1)=2/3`

Options

A. `y'(1)=4//3`
B. `y''(1)=-4//3`
C. `y''(1)=-8(22)/(27)`
D. `y'(1)=2//3`

Explanation

`x^(3)-2x^(2)y^(2)+5x+y-5=0`
Differentiating w.r.t. x, we get
`3x^(2)-4xy^(2)-4x^(2)y(dy)/(dx)+5+(dy)/(dx)=0`
`"or "y'=(dy)/(dx)=(3x^(2)-4xy^(2)+5)/(4x^(2)y-1)`
`y'(1)=(3-4+5)/(4-3)=(4)/(1)`
Also, y''
`=((6x-4y^(2)-8xyy')(4x^(2)y-1)-(8xy+4x^(2)y')(3x^(2)-4xy^(2)+5))/((4x^(2)y-1)^(2))`
`"or "y''(1)=((6-4-8xx(4)/(3))(4-1)-(8+4xx(4)/(3))(3-4+5))/((4-1)^(2))`
`=-8(22)/(27)`