Answer:
[tex]\displaystyle y' = \frac{\sqrt{y} + y}{4\sqrt{x}y \Big( y^\Big{\frac{3}{2}} + \sqrt{y} + 2y \Big) + \sqrt{x} + x}[/tex]
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Quotient Rule]: [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Implicit Differentiation
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \frac{\sqrt{x} + 1}{\sqrt{y} + 1} = y^2[/tex]
Step 2: Differentiate
- Implicit Differentiation: [tex]\displaystyle \frac{dy}{dx} \bigg[ \frac{\sqrt{x} + 1}{\sqrt{y} + 1} \bigg] = \frac{dy}{dx}[ y^2][/tex]
- Quotient Rule: [tex]\displaystyle \frac{(\sqrt{x} + 1)'(\sqrt{y} + 1) - (\sqrt{y} + 1)'(\sqrt{x} + 1)}{(\sqrt{y} + 1)^2} = \frac{dy}{dx}[ y^2][/tex]
- Rewrite: [tex]\displaystyle \frac{(x^\Big{\frac{1}{2}} + 1)'(y^\Big{\frac{1}{2}} + 1) - (y^\Big{\frac{1}{2}} + 1)'(x^\Big{\frac{1}{2}} + 1)}{(y^\Big{\frac{1}{2}} + 1)^2} = \frac{dy}{dx}[ y^2][/tex]
- Basic Power Rule [Addition/Subtraction, Chain Rule]: [tex]\displaystyle \frac{\frac{1}{2}x^\Big{\frac{-1}{2}}(y^\Big{\frac{1}{2}} + 1) - \frac{1}{2}y^\Big{\frac{-1}{2}}y'(x^\Big{\frac{1}{2}} + 1)}{(y^\Big{\frac{1}{2}} + 1)^2} = 2yy'[/tex]
- Factor: [tex]\displaystyle \frac{\frac{1}{2} \bigg[ x^\Big{\frac{-1}{2}}(y^\Big{\frac{1}{2}} + 1) - y^\Big{\frac{-1}{2}}y'(x^\Big{\frac{1}{2}} + 1) \bigg] }{(y^\Big{\frac{1}{2}} + 1)^2} = 2yy'[/tex]
- Rewrite: [tex]\displaystyle \frac{x^\Big{\frac{-1}{2}}(y^\Big{\frac{1}{2}} + 1) - y^\Big{\frac{-1}{2}}y'(x^\Big{\frac{1}{2}} + 1)}{2(y^\Big{\frac{1}{2}} + 1)^2} = 2yy'[/tex]
- Rewrite: [tex]\displaystyle x^\Big{\frac{-1}{2}}(y^\Big{\frac{1}{2}} + 1) - y^\Big{\frac{-1}{2}}y'(x^\Big{\frac{1}{2}} + 1)}= 4yy'(y^\Big{\frac{1}{2}} + 1)^2[/tex]
- Isolate y' terms: [tex]\displaystyle x^\Big{\frac{-1}{2}}(y^\Big{\frac{1}{2}} + 1) = 4yy'(y^\Big{\frac{1}{2}} + 1)^2 + y^\Big{\frac{-1}{2}}y'(x^\Big{\frac{1}{2}} + 1)}[/tex]
- Factor: [tex]\displaystyle x^\Big{\frac{-1}{2}}(y^\Big{\frac{1}{2}} + 1) = y' \bigg[ 4y(y^\Big{\frac{1}{2}} + 1)^2 + y^\Big{\frac{-1}{2}}(x^\Big{\frac{1}{2}} + 1)} \bigg][/tex]
- Isolate y': [tex]\displaystyle \frac{x^\Big{\frac{-1}{2}}(y^\Big{\frac{1}{2}} + 1)}{4y(y^\Big{\frac{1}{2}} + 1)^2 + y^\Big{\frac{-1}{2}}(x^\Big{\frac{1}{2}} + 1)} = y'[/tex]
- Rewrite/Simplify: [tex]\displaystyle y' = \frac{\sqrt{y} + y}{4\sqrt{x}y \Big( y^\Big{\frac{3}{2}} + \sqrt{y} + 2y \Big) + \sqrt{x} + x}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
Book: College Calculus 10e
