To find [tex]\(\frac{dy}{dx}\)[/tex] for the equation [tex]\(x^5 + 4xy^3 - 3y^5 = 2\)[/tex], we will use the method of implicit differentiation. Here is a step-by-step solution:
1. Differentiate both sides of the equation with respect to [tex]\(x\)[/tex]:
Given the equation:
[tex]\[ x^5 + 4xy^3 - 3y^5 = 2 \][/tex]
First, differentiate term by term taking into account that [tex]\(y\)[/tex] is a function of [tex]\(x\)[/tex]:
- The derivative of [tex]\(x^5\)[/tex] with respect to [tex]\(x\)[/tex] is:
[tex]\[ \frac{d}{dx}(x^5) = 5x^4 \][/tex]
- For the term [tex]\(4xy^3\)[/tex], we apply the product rule:
[tex]\[ \frac{d}{dx}(4xy^3) = 4 \left( x \frac{d}{dx}(y^3) + y^3 \frac{d}{dx}(x) \right) \][/tex]
[tex]\[ = 4 \left( x \cdot 3y^2 \frac{dy}{dx} + y^3 \cdot 1 \right) \][/tex]
[tex]\[ = 4 \left( 3xy^2 \frac{dy}{dx} + y^3 \right) \][/tex]
[tex]\[ = 12xy^2 \frac{dy}{dx} + 4y^3 \][/tex]
- The derivative of [tex]\(-3y^5\)[/tex] requires the chain rule:
[tex]\[ \frac{d}{dx}(-3y^5) = -3 \cdot 5y^4 \frac{dy}{dx} \][/tex]
[tex]\[ = -15y^4 \frac{dy}{dx} \][/tex]
- The derivative of the constant [tex]\(2\)[/tex] is [tex]\(0\)[/tex]:
[tex]\[ \frac{d}{dx}(2) = 0 \][/tex]
So, the differentiated equation is:
[tex]\[ 5x^4 + 12xy^2 \frac{dy}{dx} + 4y^3 - 15y^4 \frac{dy}{dx} = 0 \][/tex]
2. Isolate [tex]\(\frac{dy}{dx}\)[/tex]:
Combine the terms involving [tex]\(\frac{dy}{dx}\)[/tex] on one side of the equation:
[tex]\[ 12xy^2 \frac{dy}{dx} - 15y^4 \frac{dy}{dx} = -5x^4 - 4y^3 \][/tex]
Factor out [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \frac{dy}{dx} (12xy^2 - 15y^4) = -5x^4 - 4y^3 \][/tex]
Solve for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{-5x^4 - 4y^3}{12xy^2 - 15y^4} \][/tex]
Therefore, the derivative [tex]\(\frac{dy}{dx}\)[/tex] given the equation [tex]\(x^5 + 4xy^3 - 3y^5 = 2\)[/tex] is:
[tex]\[ \frac{dy}{dx} = \frac{5x^4 + 4y^3}{12xy^2 - 15y^4} \][/tex]
This is the slope of [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex] at any point satisfying the given equation.
