Q10P In Problem 10, find Y"at the ori... [FREE SOLUTION]

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Chapter 4: Q10P (page 203)

In Problem 10, find $Y "$ at the origin.

Short Answer

Expert verified

The second derivative of the equation $x e^{y} + y e^{x} = 0$

Is ${(\frac{d^{2} y}{d x^{2}})}_{(0, 0)} = 4$ .

Step by step solution

Explanation of Solution

The provided equation or curve is $x e^{y} + y e^{x} = 0 .$

Step 2: Implicit Differentiation

To find the slopes of tangents to curves that aren't functions, we can utilize implicit differentiation (they fail the vertical line test). Parts ofyare assumed to be functions that satisfy the supplied equation, but yis not a function ofx.

Calculation

Consider the equation of the curve below:

$x e^{y} + y e^{x} = 0$

Differentiate the provided above equation of curve with respect to x,

$x e^{y} \frac{d y}{d x} + e^{x} + y e^{x} + e^{x} \frac{d y}{d x} = 0 (x e^{y} + e^{x}) \frac{d y}{d x} + e^{x} + y e^{x} = 0$

Hence,

$\frac{d y}{d x} = - (\frac{e^{x} + y e^{x}}{x e^{y} + e^{x}})$

Again, differentiate the provided above equation of a curve,

$\frac{d^{2} y}{d x^{2}} = - (e^{x} + y e^{x}) = \frac{\{\begin{matrix} (x e^{y} + e^{x}) (e^{y} (- (\frac{e^{y} + y e^{x}}{x e^{y} + e^{x}}))) + y e^{x} + e^{x} (- (\frac{e^{y} + y e^{x}}{x e^{y} + e^{x}})) \\ - (e^{y} + y e^{x}) (x e^{y} (- (\frac{e^{y} + y e^{x}}{x e^{y} + e^{x}})) + e^{y} + e^{x}) \end{matrix}\}}{x e^{y} + e^{x}}$

At the point (0,0),

${(\frac{d^{2} y}{d x^{2}})}_{(0, 0)} = 4$

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91影视

In Problem 10, find $Y "$ at the origin.

Short Answer

Step by step solution

Explanation of Solution

Step 2: Implicit Differentiation

Calculation

One App. One Place for Learning.

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91影视

In Problem 10, find Y"at the origin.

Short Answer

Step by step solution

Explanation of Solution

Step 2: Implicit Differentiation

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Conservation of Energy and Momentum

Astrophysics

Further Mechanics and Thermal Physics

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Circular Motion and Gravitation

Study anywhere. Anytime. Across all devices.

In Problem 10, find $Y "$ at the origin.