Question

Second derivative of x^½ + y^½ = a^½

Answer 1

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

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:

*Note:

Allow a to be defined as an arbitrary constant.

Step 1: Define

`\displaystyle x^\big{(1)/(2)} + y^\big{(1)/(2)} = a^\big{(1)/(2)}`

Step 2: Differentiate

1. Basic Power Rule [Addition/Subtraction, Chain Rule]:
   `\displaystyle (1)/(2√(x)) + (y')/(2√(y)) = 0`
2. Isolate y' term:
   `\displaystyle (y')/(2√(y)) = -(1)/(2√(x))`
3. Isolate y':
   `\displaystyle y' = -(√(y))/(√(x))`
4. Derivative Rule [Quotient Rule]:
   `\displaystyle y'' = -((√(y))'(√(x)) - √(y)(√(x))')/((√(x))^2)`
5. Basic Power Rule [Derivative Rule - Chain Rule]:
   `\displaystyle y'' = -((y'√(x))/(2√(y)) - (√(y))/(2√(x)))/((√(x))^2)`
6. Simplify:
   `\displaystyle y'' = -((y'√(x))/(2√(y)) - (√(y))/(2√(x)))/(x)`
7. Rewrite:
   `\displaystyle y'' = \frac{-(y'x - y)}{2x^\big{(3)/(2)}√(y)}`
8. Substitute in y':
   `\displaystyle y'' = \frac{-(-(√(y))/(√(x))x - y)}{2x^\big{(3)/(2)}√(y)}`
9. Simplify:
   `\displaystyle y'' = \frac{√(x) + √(y)}{2x^\big{(3)/(2)}}`

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

Unit: Differentiation