   Chapter 2.6, Problem 40E

Chapter
Section
Textbook Problem

If x 2 + x y + y 3 = 1 , find the value of y ″ at the point where x = 1 .

To determine

To find:

y''' at x=1

Explanation

1) Concept:

To find y''', first differentiate the given equation with respect to x. Then solve it for y.

Again differentiate the equation of y with respect to x to get y again differentiate it with respect to x to get y implicitly, and then plug x = 1 to find value of y at x=1

2) Formula:

i. Constant Multiple rule

ddx cfx=  cddxfx

ii. Sum or difference rule

ddxfx ±gx=ddx fx± ddxgx

iii. The power rule

ddxxn=n xn-1

iv. Product rule

ddxf*g=fddxg+gddxf

3) Given:

x2+xy+y3=1

4) Calculation:

Consider the given equation

x2+xy+y3=1

At x = 1

1+1*y+y3=1

1+y+y3=1

y+y3=1-1

y+y3=0

y(1+y2)=0

y=0,±i

Consider real value

that is y=0,

When x = 1, y =0.

Now differentiate given equation with respect to x

ddxx2+xy+y3=1

ddxx2+ddxxy+ddxy3=ddx1

Product rule for differentiation

ddxx2+xdydx+ydxdx+ddxy3=ddx1

2x+xdydx+y+3y2dydx=0

Plug x = 1 and y = 0

2*1+1*dydx+0+0dydx=0

2+dydx=0 So at x =1

dydx=-2

Differentiate 2x+xdydx+y+3y2dydx=0

with respect to x

ddx2x+xdydx+y+3y2dydx=0

ddx2x+ddxxdydx+ddxy+ddx3y2dydx=0

Product rule for differentiation

ddx2x+xddxdydx+dydxddxx+ddxy+3y2ddxdydx+dydxddx3y2=0

2+xd2ydx2+dydx+dydx+3y2d2ydx2+6ydydx2=0

To find value of d2ydx2  at x=1 Plug dydx=-2, x=1 and y=0

2+1*d2ydx2+-2+-2+0*d2ydx2+6*0*dydx2=0

d

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