Q. 79 Each of the equations in Exercis... [FREE SOLUTION]

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Chapter 2: Q. 79 (page 211)

Each of the equations in Exercises 69鈥�80 defines y as an implicit function of x. Use implicit differentiation (without solving for y first) to find $\frac{d y}{d x}$
$\frac{1}{y} - \frac{1}{x} = \frac{x^{2}}{y + 1}$

Short Answer

Expert verified

$\frac{d y}{d x} = \frac{y^{2} (y + 1) (2 x^{3} - (y + 1))}{x^{2} ({(y + 1)}^{2} + x^{2} y^{2})}$

Step by step solution

Step 1. Given Information:

Given equation: $\frac{1}{y} - \frac{1}{x} = \frac{x^{2}}{y + 1}$

We want to find $\frac{d y}{d x}$ defines y as an implicit function of x by use implicit differentiation.

Step 2. Solution:

Differentiate both sides w.r.t. x

$\frac{d}{d x} (\frac{1}{y} - \frac{1}{x}) = \frac{d}{d x} (\frac{x^{2}}{y + 1}) \frac{d}{d x} \frac{1}{y} - \frac{d}{d x} \frac{1}{x} = \frac{d}{d x} x^{2} {(y + 1)}^{- 1} U \sin g p r o d u c t a n d c h a i n r u l e w e g e t \frac{1}{y^{2}} \frac{d y}{d x} + \frac{1}{x^{2}} = x^{2} \frac{d}{d x} {(y + 1)}^{- 1} + {(y + 1)}^{- 1} \frac{d}{d x} x^{2} \frac{1}{y^{2}} \frac{d y}{d x} + \frac{1}{x^{2}} = x^{2} 路 (- 1 {(y + 1)}^{- 2} \frac{d y}{d x}) + {(y + 1)}^{- 1} 路 2 x \frac{1}{y^{2}} \frac{d y}{d x} + \frac{1}{x^{2}} = \frac{- x^{2}}{{(y + 1)}^{2}} \frac{d y}{d x} + \frac{2 x}{(y + 1)} \frac{1}{y^{2}} \frac{d y}{d x} + \frac{x^{2}}{{(y + 1)}^{2}} \frac{d y}{d x} = \frac{2 x}{(y + 1)} - \frac{1}{x^{2}} (\frac{1}{y^{2}} + \frac{x^{2}}{{(y + 1)}^{2}}) \frac{d y}{d x} = \frac{2 x}{(y + 1)} - \frac{1}{x^{2}} (\frac{{(y + 1)}^{2} + x^{2} y^{2}}{{y^{2} (y + 1)}^{2}}) \frac{d y}{d x} = \frac{2 x^{3} - (y + 1)}{x^{2} (y + 1)} \frac{d y}{d x} = (\frac{2 x^{3} - (y + 1)}{x^{2} (y + 1)}) (\frac{{y^{2} (y + 1)}^{2}}{{(y + 1)}^{2} + x^{2} y^{2}}) \frac{d y}{d x} = \frac{y^{2} (y + 1) (2 x^{3} - (y + 1))}{x^{2} ({(y + 1)}^{2} + x^{2} y^{2})}$

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91影视

Each of the equations in Exercises 69鈥�80 defines y as an implicit function of x. Use implicit differentiation (without solving for y first) to find $\frac{d y}{d x}$
$\frac{1}{y} - \frac{1}{x} = \frac{x^{2}}{y + 1}$

Short Answer

Step by step solution

Step 1. Given Information:

Step 2. Solution:

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Each of the equations in Exercises 69鈥�80 defines y as an implicit function of x. Use implicit differentiation (without solving for y first) to find dydx1y-1x=x2y+1

Short Answer

Step by step solution

Step 1. Given Information:

Step 2. Solution:

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Theoretical and Mathematical Physics

Decision Maths

Pure Maths

Mechanics Maths

Applied Mathematics

Study anywhere. Anytime. Across all devices.

Each of the equations in Exercises 69鈥�80 defines y as an implicit function of x. Use implicit differentiation (without solving for y first) to find $\frac{d y}{d x}$
$\frac{1}{y} - \frac{1}{x} = \frac{x^{2}}{y + 1}$