`(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)`

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

(y-x(dy)/(dx))=a(y^(2)+(dy)/(dx))

(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

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))

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)

(dy)/(dx)=(y^(2)-y+1)/(x^(2)-x+1)

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

If e^y(x+1)=1 , prove that (d^2y)/(dx^2)=((dy)/(dx))^2

If (a+bx)e^((y)/(x))=x , show that, x^(3)(d^(2)y)/(dx^(2))=(x(dy)/(dx)-y)^(2) .