Q. 80 Each of the equations in Exercis... [FREE SOLUTION]

91影视

Chapter 2: Q. 80 (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 $\frac{d y}{d x}$
$\frac{x + 1}{y^{2} - 3} = \frac{1}{x y}$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information:

Given equation: $\frac{x + 1}{y^{2} - 3} = \frac{1}{x y}$

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

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

Using product and chain rule we get

localid="1648704342113" $(x + 1) \frac{d}{d x} {(y^{2} - 3)}^{- 1} + {(y^{2} - 3)}^{- 1} \frac{d}{d x} (x + 1) = \frac{d}{d x} {(x y)}^{- 1} (x + 1) ({- 1 (y^{2} - 3)}^{- 2} \frac{d}{d x} (y^{2} - 3)) + {(y^{2} - 3)}^{- 1} \frac{d}{d x} (x + 1) = {(- 1) (x y)}^{- 2} \frac{d}{d x} x y (x + 1) (\frac{- 1}{{(y^{2} - 3)}^{2}} (2 y \frac{d y}{d x} - 0)) + \frac{1}{(y^{2} - 3)} (1) = \frac{- 1}{{(x y)}^{2}} (x \frac{d y}{d x} + y) \frac{- 2 y (x + 1)}{{(y^{2} - 3)}^{2}} \frac{d y}{d x} + \frac{1}{(y^{2} - 3)} = \frac{- 1}{{x y}^{2}} \frac{d y}{d x} - \frac{1}{x^{2} y} \frac{- 1}{{x y}^{2}} \frac{d y}{d x} + \frac{2 y (x + 1)}{{(y^{2} - 3)}^{2}} \frac{d y}{d x} = \frac{1}{(y^{2} - 3)} + \frac{1}{x^{2} y} (\frac{- 1}{{x y}^{2}} + \frac{2 y (x + 1)}{{(y^{2} - 3)}^{2}}) \frac{d y}{d x} = \frac{1}{(y^{2} - 3)} + \frac{1}{x^{2} y} (\frac{- {(y^{2} - 3)}^{2} + 2 x y^{3} (x + 1)}{x y^{2} {(y^{2} - 3)}^{2}}) \frac{d y}{d x} = \frac{x^{2} y + (y^{2} - 3)}{x^{2} y (y^{2} - 3)} \frac{d y}{d x} = (\frac{x^{2} y + (y^{2} - 3)}{x^{2} y (y^{2} - 3)}) (\frac{x y^{2} {(y^{2} - 3)}^{2}}{2 x y^{3} (x + 1) - {(y^{2} - 3)}^{2}}) \frac{d y}{d x} = \frac{y {(y^{2} - 3)}^{2} (x^{2} y + (y^{2} - 3))}{x (2 x y^{3} (x + 1) - {(y^{2} - 3)}^{2})}$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

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

Short Answer

Step by step solution

Step 1. Given Information:

Step 2. Solution:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Theoretical and Mathematical Physics

Pure Maths

Mechanics Maths

Discrete Mathematics

Probability and Statistics

Study anywhere. Anytime. Across all devices.

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 dydxx+1y2-3=1xy

Short Answer

Step by step solution

Step 1. Given Information:

Step 2. Solution:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Theoretical and Mathematical Physics

Pure Maths

Mechanics Maths

Discrete Mathematics

Probability and Statistics

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