Chapter 2: Q. 77 (page 211)

Each of the equations in Exercises 69–80 defines y as an implicit function of x. Use implicit differentiation (without solving for y first) to find $\frac{d y}{d x}$
$\frac{y^{2} + 1}{3 y - 1} = x$

Short Answer

Expert verified

$\frac{d y}{d x} = \frac{- 1}{3} \cdot \frac{{(3 y - 1)}^{2}}{7 y^{2} - 2 y + 1}$

Step by step solution

Step 1. Given Information:

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

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

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

Using product and chain rule we get

$(y^{2} + 1) \frac{d}{d x} {(3 y - 1)}^{- 1} + {(3 y - 1)}^{- 1} \frac{d}{d x} (y^{2} + 1) = \frac{d}{d x} x (y^{2} + 1) ({- 1 (3 y - 1)}^{- 2}) \frac{d}{d x} (3 y - 1) + {(3 y - 1)}^{- 1} (2 y \frac{d}{d x} y + 0) = \frac{d}{d x} x (y^{2} + 1) ({- 1 (3 y - 1)}^{- 2}) (3 \frac{d y}{d x} - 0) + {(3 y - 1)}^{- 1} (2 y \frac{d y}{d x}) = 1 (- 3) \frac{y^{2} + 1}{{(3 y - 1)}^{2}} \frac{d y}{d x} + \frac{2 y}{3 y - 1} \frac{d y}{d x} = 1 ((- 3) \frac{y^{2} + 1}{{(3 y - 1)}^{2}} + \frac{2 y}{3 y - 1}) \frac{d y}{d x} = 1 ((- 3) \frac{y^{2} + 1 + 2 y (3 y - 1)}{{(3 y - 1)}^{2}}) \frac{d y}{d x} = 1 ((- 3) \frac{y^{2} + 1 + 6 y^{2} - 2 y}{{(3 y - 1)}^{2}}) \frac{d y}{d x} = 1 ((- 3) \frac{7 y^{2} - 2 y + 1}{{(3 y - 1)}^{2}}) \frac{d y}{d x} = 1 \frac{d y}{d x} = \frac{- 1}{3} \cdot \frac{{(3 y - 1)}^{2}}{7 y^{2} - 2 y + 1}$

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Each of the equations in Exercises 69–80 defines y as an implicit function of x. Use implicit differentiation (without solving for y first) to find $\frac{d y}{d x}$
$\frac{y^{2} + 1}{3 y - 1} = x$

Short Answer

Step by step solution

Step 1. Given Information:

Step 2. Solution:

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Each of the equations in Exercises 69–80 defines y as an implicit function of x. Use implicit differentiation (without solving for y first) to find dydxy2+13y-1=x

Short Answer

Step by step solution

Step 1. Given Information:

Step 2. Solution:

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Each of the equations in Exercises 69–80 defines y as an implicit function of x. Use implicit differentiation (without solving for y first) to find $\frac{d y}{d x}$
$\frac{y^{2} + 1}{3 y - 1} = x$