# The derivative d x d y by using 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 19E
To determine

## To find: The derivative dxdy by using implicit differentiation.

Expert Solution

The implicit differentiation is dxdy=2x4y+x36xy24x3y23x2y+2y3.

### Explanation of Solution

Given:

The equation is x4y2x3y+2xy3=0.

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(f(x)g(x))=f(x)ddx(g(x))+g(x)ddx(f(x)).

Calculation:

Let the independent variable be y and the dependent variable be x.

x4y2x3y+2xy3=0

Differentiate implicitly with respect to y on both sides,

ddy(x4y2x3y+2xy3)=ddy(0)ddy(x4y2)ddy(x3y)+ddy(2xy3)=0ddy(x4y2)ddy(x3y)+2ddy(xy3)=0

Apply the product rule,

[x4ddy(y2)+y2ddy(x4)][x3ddy(y)+yddy(x3)]+2[xddy(y3)+y3ddy(x)]=0[x4(2y)+y2ddy(x4)][x3(1)+yddy(x3)]+2[x(3y2)+y3ddy(x)]=0

Apply the chain rule (1),

[x4(2y)+y2[ddx(x4)dxdy]][x3(1)+y[ddx(x3)dxdy]]+2[x(3y2)+y3dxdy]=0[x4(2y)+y2[4x3dxdy]][x3(1)+y[3x2dxdy]]+2[x(3y2)+y3dxdy]=0[2x4y+4x3y2dxdy][x3+3x2ydxdy]+2[3xy2+y3dxdy]=0[2x4y+4x3y2dxdy][x3+3x2ydxdy]+[6xy2+2y3dxdy]=0

Simplify further and obtain the derivative.

2x4y+4x3y2dxdyx33x2ydxdy+6xy2+2y3dxdy=0dxdy(4x3y23x2y+2y3)=2x4y+x36xy2dxdy=2x4y+x36xy24x3y23x2y+2y3

Therefore, the implicit differentiation is dxdy=2x4y+x36xy24x3y23x2y+2y3.

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