Find (dy)/(dx) if y = (x + √(x) )^(2) ​

Question

Find

`(dy)/(dx)`
if

`y = (x + √(x) )^(2)`
​

Answer 1

`\displaystyle y' = 2x + 3√(x) + 1`

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Algebra I

• Terms/Coefficients
• Anything to the 0th power is 1
• Exponential Rule [Rewrite]:
  `\displaystyle b^(-m) = (1)/(b^m)`
• Exponential Rule [Root Rewrite]:
  `\displaystyle \sqrt[n]{x} = x^{(1)/(n)}`

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

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

Explanation:

Step 1: Define

Identify

`\displaystyle y = (x + √(x))^2`

Step 2: Differentiate

1. Chain Rule:
   `\displaystyle y' = 2(x + √(x))^(2 - 1) \cdot (d)/(dx)[x + √(x)]`
2. Rewrite [Exponential Rule - Root Rewrite]:
   `\displaystyle y' = 2(x + x^{(1)/(2)})^(2 - 1) \cdot (d)/(dx)[x + x^{(1)/(2)}]`
3. Simplify:
   `\displaystyle y' = 2(x + x^{(1)/(2)}) \cdot (d)/(dx)[x + x^{(1)/(2)}]`
4. Basic Power Rule:
   `\displaystyle y' = 2(x + x^{(1)/(2)}) \cdot (1 \cdot x^(1 - 1) + (1)/(2)x^{(1)/(2) - 1})`
5. Simplify:
   `\displaystyle y' = 2(x + x^{(1)/(2)}) \cdot (1 + (1)/(2)x^{-(1)/(2)})`
6. Rewrite [Exponential Rule - Rewrite]:
   `\displaystyle y' = 2(x + x^{(1)/(2)}) \cdot (1 + \frac{1}{2x^{(1)/(2)}})`
7. Multiply:
   `\displaystyle y' = 2[(x + x^{(1)/(2)}) + \frac{x + x^{(1)/(2)}}{2x^{(1)/(2)}}]`
8. [Brackets] Add:
   `\displaystyle y' = 2(\frac{2x + 3x^{(1)/(2)} + 1}{2})`
9. Multiply:
   `\displaystyle y' = 2x + 3x^{(1)/(2)} + 1`
10. Rewrite [Exponential Rule - Root Rewrite]:
    `\displaystyle y' = 2x + 3√(x) + 1`

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

Unit: Derivatives

Book: College Calculus 10e