The value of y ′ ′ ′ at the point x = 1 .

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.5, Problem 36E
To determine

To find: The value of y′′′ at the point x=1.

Expert Solution

The derivative of the equation at (1,0) is y(x)=42.

Explanation of Solution

Given:

The equation x2+xy+y3=1.

Derivative rules:

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

(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)).

Calculation:

The value of y if x=1 is computed as follows,

Substitute the value x=1 in x2+xy+y3=1,

12+(1)y+y3=1y3+y=0y(y2+1)=0y=0 or y=±i

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

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

x2+xy+y3=1

Differentiate implicitly with respect to x,

ddx(x2+xy+y3)=ddx(1)ddx(x2)+ddx(xy)+ddx(y3)=02x+ddx(xy)+ddx(y3)=0

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

2x+[xddx(y)+yddx(x)]+[ddy(y3)dydx]=02x+[xdydx+y(1)]+[3y2dydx]=0

2x+xdydx+y+3y2dydx=0 (1)

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

2(1)+(1)dydx+0+3(0)2dydx=02+dydx=0dydx=2

Thus, the derivative y at (1,0)y=2.

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

ddx(2x+xdydx+y+3y2dydx)=ddx(0)ddx(2x)+ddx(xy)+ddx(y)+ddx(3y2y)=02ddx(x)+ddx(xy)+ddx(y)+3ddx(y2y)=02+ddx(xy)+3ddx(y2y)+dydx=0

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

2+[xddx(y)+yddx(x)]+3[y2ddx(y)+yddx(y2)]+dydx=02+[xy+y(1)]+3[y2y+y(ddy(y2)dydx)]+y=02+xy+y+3[y2y+y(2yy)]+y=0

2+xy+2y+3y2y+6y(y)2=0 (2)

Substitute (1,0) for (x,y) and y=2 in (1),

2+(1)y+2(2)+3(0)y+6(0)(2)2=02+y4=0y2=0y=2

Thus, the derivative y at (1,0) y=2.

Differentiate equation (2) implicitly with respect to x,

ddx(2+xy+2y+3y2y+6y(y)2)=ddx(0)0+ddx(xy)+2y+3[ddx(y2y)]+6[ddx(y(y)2)]=0ddx(xy)+2y+3[ddx(y2y)]+6[ddx(y(y)2)]=0

Apply the product rule (2) and simplify further,

{[xddx(y)+yddx(x)]+2y+3[y2ddx(y)+yddx(y2)]+6[yddx((y)2)+(y)2ddx(y)]}=0[xy+y(1)]+2y+3[y2(y)+y2y]+6[y(ddy((y)2)ddx(y))+(y)2y]=0[xy+y]+2y+3[y2(y)+y2y]+6[y(2yy)+(y)2y]=0

Substitute (1,0) for (x,y), y=2 and y=2 in equation (2),

[(1)y+2]+2(2)+3[02(y)+(2)2(0)]+6[(0)(2(2)(2))+(2)2(2)]=0y+2+4+6(8)=0y=42

Therefore, the derivative of the equation at (1,0) is y(x)=42.

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!