Question

Find y' for the following. ​

Answer 1

`\displaystyle y' = \frac{√(y) + y}{4√(x)y \Big( y^\Big{(3)/(2)} + √(y) + 2y \Big) + √(x) + x}`

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:
`\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)`

Derivative Property [Addition/Subtraction]:
`\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]`

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]:
`\displaystyle (d)/(dx) [(f(x))/(g(x)) ]=(g(x)f'(x)-g'(x)f(x))/(g^2(x))`

Derivative Rule [Chain Rule]:
`\displaystyle (d)/(dx)[f(g(x))] =f'(g(x)) \cdot g'(x)`

Implicit Differentiation

Explanation:

Step 1: Define

Identify

`\displaystyle (√(x) + 1)/(√(y) + 1) = y^2`

Step 2: Differentiate

1. Implicit Differentiation:
   `\displaystyle (dy)/(dx) \bigg[ (√(x) + 1)/(√(y) + 1) \bigg] = (dy)/(dx)[ y^2]`
2. Quotient Rule:
   `\displaystyle ((√(x) + 1)'(√(y) + 1) - (√(y) + 1)'(√(x) + 1))/((√(y) + 1)^2) = (dy)/(dx)[ y^2]`
3. Rewrite:
   `\displaystyle \frac{(x^\Big{(1)/(2)} + 1)'(y^\Big{(1)/(2)} + 1) - (y^\Big{(1)/(2)} + 1)'(x^\Big{(1)/(2)} + 1)}{(y^\Big{(1)/(2)} + 1)^2} = (dy)/(dx)[ y^2]`
4. Basic Power Rule [Addition/Subtraction, Chain Rule]:
   `\displaystyle \frac{(1)/(2)x^\Big{(-1)/(2)}(y^\Big{(1)/(2)} + 1) - (1)/(2)y^\Big{(-1)/(2)}y'(x^\Big{(1)/(2)} + 1)}{(y^\Big{(1)/(2)} + 1)^2} = 2yy'`
5. Factor:
   `\displaystyle \frac{(1)/(2) \bigg[ x^\Big{(-1)/(2)}(y^\Big{(1)/(2)} + 1) - y^\Big{(-1)/(2)}y'(x^\Big{(1)/(2)} + 1) \bigg] }{(y^\Big{(1)/(2)} + 1)^2} = 2yy'`
6. Rewrite:
   `\displaystyle \frac{x^\Big{(-1)/(2)}(y^\Big{(1)/(2)} + 1) - y^\Big{(-1)/(2)}y'(x^\Big{(1)/(2)} + 1)}{2(y^\Big{(1)/(2)} + 1)^2} = 2yy'`
7. Rewrite:
   `\displaystyle x^\Big{(-1)/(2)}(y^\Big{(1)/(2)} + 1) - y^\Big{(-1)/(2)}y'(x^\Big{(1)/(2)} + 1)}= 4yy'(y^\Big{(1)/(2)} + 1)^2`
8. Isolate y' terms:
   `\displaystyle x^\Big{(-1)/(2)}(y^\Big{(1)/(2)} + 1) = 4yy'(y^\Big{(1)/(2)} + 1)^2 + y^\Big{(-1)/(2)}y'(x^\Big{(1)/(2)} + 1)}`
9. Factor:
   `\displaystyle x^\Big{(-1)/(2)}(y^\Big{(1)/(2)} + 1) = y' \bigg[ 4y(y^\Big{(1)/(2)} + 1)^2 + y^\Big{(-1)/(2)}(x^\Big{(1)/(2)} + 1)} \bigg]`
10. Isolate y':
    `\displaystyle \frac{x^\Big{(-1)/(2)}(y^\Big{(1)/(2)} + 1)}{4y(y^\Big{(1)/(2)} + 1)^2 + y^\Big{(-1)/(2)}(x^\Big{(1)/(2)} + 1)} = y'`
11. Rewrite/Simplify:
    `\displaystyle y' = \frac{√(y) + y}{4√(x)y \Big( y^\Big{(3)/(2)} + √(y) + 2y \Big) + √(x) + x}`

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Book: College Calculus 10e