If `R=([1+((dy)/(dx))^2]^(3//2))/((d^2y)/(dx2))` , then`R^(2//3)` can be put in the form of a. `1/(((d^2y)/(dx^2))^(2//3))+1/(((d^2x)/(dy^2))^(2//3))` b. `1/(((d^2y)/(dx^2))^(2//3))-1/(((d^2x)/(dy^2))^(2//3))` c. `2/(((d^2y)/(dx^2))^(2//3))+2/(((d^2x)/(dy^2))^(2//3))` d. `1/(((d^2y)/(dx^2))^(2//3))1/(((d^2x)/(dy^2))^(2//3))`

(d^2x)/(dy^2) equals: (1) ((d^2y)/(dx^2))^(-1) (2) -((d^2y)/(dx^2))^(-1)((dy)/(dx))^(-3) (3) ((d^2y)/(dx^2))^(-1)((dy)/(dx))^(-2) (4) -((d^2y)/(dx^2))((dy)/(dx))^(-3)

(dy)/(dx)=(3x+2y)/(2x-3y)

If y^(2)=4ax , then the value of (d^(2)y)/(dx^(2))*(d^(2)x)/(dy^(2)) is -

(dy)/(dx)=(2x-3y)/(3x-2y)

(dy)/(dx)=(2x+3y)/(3x+2y)

If x=logp and y=1/p ,then (a) (d^2y)/(dx^2)-2p=0 (b) (d^2y)/(dx^2)+y=0 (c) (d^2y)/(dx^2)+(dy)/(dx)=0 (d) (d^2y)/(dx^2)-(dy)/(dx)=0

In the parabola y^2=4ax , prove that (d^2y)/(dx^2).(d^2x)/(dy^2)=(-2a)/y^3

Solve (d^2y)/(dx^2)=((dy)/(dx))^2

The second derivative of a single valued function parametrically represented by x=varphi(t)a n dy=psi(t) (where varphi(t)a n dpsi(t) are different function and varphi^(prime)(t)!=0 ) is given by (d^2y)/(dx^2)=(((dx)/(dt))((d^2y)/(dt^2))-((d^2x)/(dt^2))((dy)/(dt))/(((dx)/(dt))^2) (d^2y)/(dx^2)=(((dx)/(dt))((d^2y)/(dt^2))-((d^2x)/(dt^2))((dy)/(dt))/(((dx)/(dt))^3) (d^2y)/(dx^2)=(((d^2x)/(dt))((d^y)/(dt^2))-((d^x)/(dt^))((d^2y)/(dt^2))/(((dx)/(dt))^3) (d^2y)/(dx^2)=(((d^2x)/(dt^2))((d^y)/(dt^))-((d^2x)/(dt^2))((d^y)/(dt^))/(((dx)/(dt))^3)

The degree of the differential equation ((d^(2) y)/(dx^(2)))^(2) + (d^(2)y)/(dx^(2)) - ((dy)/(dx))^(4) + (dy)/(dx) + y = 6x^(3) is -