To determine

To find: The value of y′′ at the point x=0.

Answer

The second derivative of the equation at (0,1) is y′′=0.

Explanation

Given:

The equation xy+y3=1.

Derivative rules:

(1) Chain rule: If y=f(u) and u=g(x)  are both differentiable function, then dydx=dydu⋅dudx.

(2) Product Rule: If f1(x) and f2(x) are both differentiable, then ddx(f1(x)f2(x))=f1(x)ddx(f2(x))+f2(x)ddx(f1(x)).

(3). Power Rule: ddx(xn)=nxn−1

(4) Sum Rule: ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(x)]

Calculation:

The value of y if x=0 is computed as follows,

Substitute the value x=0 in xy+y3=1,

(0)y+y3=1y3=1y=1

Thus, the point is (x,y)=(0,1).

Obtain the second derivative of the equation at (0,1).

xy+y3=1

Differentiate implicitly with respect to x,

ddx(xy+y3)=ddx(1)ddx(xy)+ddx(y3)=0

Apply the product rule (2) and the power rule (3),

[xddx(y)+yddx(x)]+ddx(y3)=0[xdydx+y(1)]+3y3−1dydx=0

[xdydx+y]+3y2dydx=0 (1)

Substitute (0,1) for (x,y) in equation (1),

[(0)dydx+(1)]+3(1)2dydx=01+3dydx=03dydx=−1dydx=−13

Thus, the derivative y′ at (0,1) is y′=−13.

Differentiate the equation (1) implicitly with respect to x,

ddx(xy′+y+3y2y′)=ddx(0)ddx(xy′)+ddx(y)+ddx(3y2y′)=0

Apply the product rule (2) and the chain rule (1),

[xddx(y′)+y′ddx(x)]+ddx(y)+[3y2ddx(y′)+y′ddx(3y2)]=0[xy′′+y′(1)]+dydx+[3y2y′′+y′(6y2−1)dydx]=0xy′′+y′+y′+3y2y′′+6yy′y′=0y′′(x+3y2)+y′(2+6yy′)=0

Substitute (0,1) for (x,y) and y′=−13,

y′′(0+3(1)2)+(−13)(2+6(1)(−13))=03y′′−13(2−2)=03y′′=0y′′=0

Therefore, the second derivative of the equation at the point (0,1) is y′′=0.