The image presents a calculus problem: "Find $\frac{d^2y}{dx^2}$ in terms of $x$ and $y$."The equation given is:\[ x^2y^2 - 8x = 3 \]Below the equation, there is a box to fill in the solution for:\[ \frac{d^2y}{dx^2} = \]A red "X" mark is next to the box, indicating that this is likely an interactive exercise where users must input their answer.

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Description: The image presents a calculus problem: "Find $\frac{d^2y}{dx^2}$ in terms of $x$ and $y$."The equation given is:\[ x^2y^2 - 8x = 3 \]Below the equation, there is a box to fill in the solution for:\[ \frac{d^2y}{dx^2} = \]A red "X" mark is next

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Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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The image presents a calculus problem:

"Find $\frac{d^2y}{dx^2}$ in terms of $x$ and $y$."

The equation given is:

\[ x^2y^2 - 8x = 3 \]

Below the equation, there is a box to fill in the solution for:

\[ \frac{d^2y}{dx^2} = \]

A red "X" mark is next to the box, indicating that this is likely an interactive exercise where users must input their answer.

Expert Solution

Step 1

Given that $x^{2} y^{2} - 8 x = 3$ .

Differentiate both sides of the above equation with respect to x as follows.

$\frac{d}{d x} (x^{2} y^{2} - 8 x) = \frac{d}{d x} (3) \frac{d}{d x} (x^{2} y^{2}) + \frac{d}{d x} (- 8 x) = 0 x^{2} \frac{d}{d x} (y^{2}) + y^{2} \frac{d}{d x} (x^{2}) - 8 = 0 x^{2} (2 y \frac{d y}{d x}) + y^{2} (2 x) = 8 2 x^{2} y \frac{d y}{d x} + 2 x y^{2} = 8 2 x y (x \frac{d y}{d x} + y) = 8 x \frac{d y}{d x} + y = \frac{4}{x y} x \frac{d y}{d x} = \frac{4}{x y} - y \frac{d y}{d x} = \frac{4 - x y^{2}}{x^{2} y}$

Thus, $\frac{d y}{d x} = \frac{4 - x y^{2}}{x^{2} y}$ .

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