Can you help me with this problem with step by step explanation, please? Thank you:) ### Using Implicit Differentiation to Find dy/dxGiven the equation: \[ x^2 + 5xy + y^2 = 0 \]We are asked to use implicit differentiation to find $\frac{dy}{dx}$.The process of implicit differentiation involves differentiating both sides of the equation with respect to $ x $, while treating $ y $ as a function of $ x $.Step-by-step differentiation:1. Differentiate $ x^2 $:\[ \frac{d}{dx}(x^2) = 2x \]2. Differentiate $ 5xy $ using the product rule ($ u \cdot v $ where $ u = 5x $ and $ v = y $):\[ \frac{d}{dx}(5xy) = 5y + 5x \frac{dy}{dx} \]3. Differentiate $ y^2 $ using the chain rule:\[ \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \]Now, combining these results, we differentiate the entire equation:\[ 2x + 5y + 5x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0 \]Next, solve for $\frac{dy}{dx}$:1. Group all the $\frac{dy}{dx}$ terms on one side:\[ 5x \frac{dy}{dx} + 2y \frac{dy}{dx} = -2x - 5y \]2. Factor out $\frac{dy}{dx}$:\[ \frac{dy}{dx}(5x + 2y) = -2x - 5y \]3. Finally, solve for $\frac{dy}{dx}$:\[ \frac{dy}{dx} = \frac{-2x - 5y}{5x + 2y} \]Thus, the implicit differentiation gives us:\[ \boxed{\frac{dy}{dx} = \frac{-2x - 5y}{5x + 2y}} \]This equation describes the slope of the tangent line to the curve $ x^2 + 5xy + y^2 = 0 $ at any point $(x, y)$.

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Use implicit differentiation to find dy/dx. x2 + 5xy + y = 0 + dy %D dx

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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Can you help me with this problem with step by step explanation, please? Thank you:)

### Using Implicit Differentiation to Find dy/dx

Given the equation:
\[ x^2 + 5xy + y^2 = 0 \]

We are asked to use implicit differentiation to find $\frac{dy}{dx}$.

The process of implicit differentiation involves differentiating both sides of the equation with respect to $ x $, while treating $ y $ as a function of $ x $.

Step-by-step differentiation:

1. Differentiate $ x^2 $:
\[ \frac{d}{dx}(x^2) = 2x \]

2. Differentiate $ 5xy $ using the product rule ($ u \cdot v $ where $ u = 5x $ and $ v = y $):
\[ \frac{d}{dx}(5xy) = 5y + 5x \frac{dy}{dx} \]

3. Differentiate $ y^2 $ using the chain rule:
\[ \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \]

Now, combining these results, we differentiate the entire equation:
\[ 2x + 5y + 5x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0 \]

Next, solve for $\frac{dy}{dx}$:

1. Group all the $\frac{dy}{dx}$ terms on one side:
\[ 5x \frac{dy}{dx} + 2y \frac{dy}{dx} = -2x - 5y \]

2. Factor out $\frac{dy}{dx}$:
\[ \frac{dy}{dx}(5x + 2y) = -2x - 5y \]

3. Finally, solve for $\frac{dy}{dx}$:
\[ \frac{dy}{dx} = \frac{-2x - 5y}{5x + 2y} \]

Thus, the implicit differentiation gives us:
\[ \boxed{\frac{dy}{dx} = \frac{-2x - 5y}{5x + 2y}} \]

This equation describes the slope of the tangent line to the curve $ x^2 + 5xy + y^2 = 0 $ at any point $(x, y)$.

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