Answer:
[tex]\displaystyle \frac{dx}{dy} = \frac{3\sqrt{\frac{3}{y}}(y - 1)}{6y}[/tex]
General Formulas and Concepts:
Algebra I
- Terms/Coefficients
- Functions
- Function Notation
Calculus
Derivatives
Derivative Notation
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Product Rule]: [tex]\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)[/tex]
Implicit Differentiation
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle y = \frac{3}{x^2}[/tex]
Step 2: Differentiate
- Rewrite Function: [tex]\displaystyle yx^2 = 3[/tex]
- [Implicit Differentiation] Basic Power Rule [Product Rule]: [tex]\displaystyle y^{1 - 1}x^2 + y(2x^{2 - 1})\frac{dx}{dy} = 3[/tex]
- Simplify: [tex]\displaystyle x^2 + y(2x)\frac{dx}{dy} = 3[/tex]
- Isolate [tex]\displaystyle \frac{dx}{dy}[/tex] term: [tex]\displaystyle y(2x)\frac{dx}{dy} = 3 - x^2[/tex]
- Isolate [tex]\displaystyle \frac{dx}{dy}[/tex]: [tex]\displaystyle \frac{dx}{dy} = \frac{3 - x^2}{2xy}[/tex]
- Substitute in x²: [tex]\displaystyle \frac{dx}{dy} = \frac{3 - \frac{3}{y}}{2xy}[/tex]
- Simplify: �� [tex]\displaystyle \frac{dx}{dy} = \frac{3(y - 1)}{2xy^2}[/tex]
- Substitute in x: [tex]\displaystyle \frac{dx}{dy} = \frac{3(y - 1)}{2y^2(\sqrt{\frac{3}{y}})}[/tex]
- Simplify: [tex]\displaystyle \frac{dx}{dy} = \frac{3\sqrt{\frac{3}{y}}(y - 1)}{6y}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Derivatives
Book: College Calculus 10e
