# The second derivative of the function x 3 + y 3 = 1 by implicit differentiation.

### 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 33E
To determine

## To calculate: The second derivative of the function x3+y3=1 by implicit differentiation.

Expert Solution

The second derivative of the function is y''=2x(y3+x2)y5 .

### Explanation of Solution

Given information:

The function x3+y3=1 .

Formula used:

Thechain rule for differentiation is if f is a function of gthen ddx(f(g(x)))=f'(g(x))g'(x) .

Power rule for differentiation is ddxxn=nxn1 .

Quotient rule for differentiation is ddx(fg)=g(x)f'(x)f(x)g'(x)[g(x)]2 where f and g are functions of x .

Calculation:

Consider the function x3+y3=1 .

Differentiate both sides with respect to x ,

ddx(x3+y3)=ddx(1)ddx(x3)+ddx(y3)=0

Recall that power rule for differentiation is ddxxn=nxn1 and chain rule for differentiation is if f is a function of gthen ddx(f(g(x)))=f'(g(x))g'(x) .

Apply it. Also observe that y is a function of x,

ddx(x3+y3)=ddx(1)ddx(x3)+ddx(y3)=03x2+3y2y'=0x2+y2y'=0

Isolate the value of y' on left hand side and simplify,

x2+y2y'=0y2y'=x2y'=x2y2

Therefore, the value of first derivative of the function is y'=x2y2 .

Now, again differentiate the above expression with respect to x . Recall that quotient rule for differentiation is ddx(fg)=g(x)f'(x)f(x)g'(x)[g(x)]2 where f and g are functions of x .

Apply it. Also observe that y is a function of x,

ddx(y')=ddx(x2y2)y''=(y2)ddx(x2)(x2)ddx(y2)(y2)2y''=y2(2x)+x2(2yy')y4y''=2xy2+2x2yy'y4

Now, substitute the value y'=x2y2 in the above expression.

y''=2xy2+2x2y(x2y2)y4=2xy+2x2(x2y)y3=2xy32x4y5=2x(y3+x2)y5

Thus, the second derivative of the function is y''=2x(y3+x2)y5 .

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