Question

Y=sqrt(x)(8x-5) find the derivative

Answer 1

`\displaystyle y' = (24x - 5)/(2√(x))`

General Formulas and Concepts:

Algebra I

• Exponentials [Fractions] - Are radicals
• Exponential Rule [Rewrite]:
  `\displaystyle b^(-m) = (1)/(b^m)`

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Product Rule:
`\displaystyle (d)/(dx) [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)`

Explanation:

Step 1: Define

`\displaystyle y = √(x)(8x - 5)`

Step 2: Differentiate

`\displaystyle f(x) = √(x), \ g(x) = (8x - 5)`

1. Product Rule:
   `\displaystyle y' = (d)/(dx)[√(x)] \cdot (8x - 5) + √(x) \cdot (d)/(dx)[(8x - 5)]`
2. Rewrite:
   `\displaystyle y' = (d)/(dx)[x^{(1)/(2)}] \cdot (8x - 5) + √(x) \cdot (d)/(dx)[(8x - 5)]`
3. Basic Power Rule:
   `\displaystyle y' = (1)/(2)x^{(1)/(2) - 1} \cdot (8x - 5) + √(x) \cdot 1 \cdot 8x^(1 - 1)`
4. Simplify:
   `\displaystyle y' = (1)/(2)x^{-(1)/(2)} \cdot (8x - 5) + √(x) \cdot 1 \cdot 8x^(0)`
5. Rewrite:
   `\displaystyle y' = \frac{1}{2x^{(1)/(2)}} \cdot (8x - 5) + √(x) \cdot 1 \cdot 8`
6. Multiply:
   `\displaystyle y' = \frac{8x + 5}{2x^{(1)/(2)}} + 8√(x)`
7. Rewrite:
   `\displaystyle y' = (8x + 5)/(2√(x)) + 8√(x)`
8. Add/Rewrite:
   `\displaystyle y' = (24x - 5)/(2√(x))`