# The value of y ′ ′ at the point x = 0 . ### 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 35E
To determine

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

Expert Solution

The second derivative of the equation at (0,1) is y=1e2.

### Explanation of Solution

Given:

The equation xy+ey=e.

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=0 is computed as follows,

Substitute the value x=0 in xy+ey=e,

(0)y+ey=eey=e

Take natural logarithmic on both sides,

lney=lney=1              (Qlne=1)

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

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

xy+ey=e

Differentiate implicitly with respect to x,

ddx(xy+ey)=ddx(e)ddx(xy)+ddx(ey)=0

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

[xddx(y)+yddx(x)]+ddx(ey)=0[xdydx+y(1)]+ddy(ey)dydx=0[xy+y]+eyy=0

xy+y+eyy=0 (1)

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

(0)y+1+e1y=00+1+ey=0ey=1y=1e

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

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

ddx(xy+y+eyy)=ddx(0)ddx(xy)+ddx(y)+ddx(eyy)=0

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

[xddx(y)+yddx(x)]+ddx(y)+[eyddx(y)+yddx(ey)]=0[xy+y(1)]+dydx+[eyy+yddy(ey)dydx]=0xy+y(1)+y+eyy+yeyy=0y(x+ey)+y(2+eyy)=0

Substitute (0,1) for (x,y) and y=1e,

y(0+e1)+1e(2+e1(1e))=0ye1e(21)=0ye=1ey=1e2

Therefore, the second derivative of the equation at the point (0,1) is y=1e2.

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