Question

Find from first principle the derivative of 3x+5/√x​

Answer 1

`\displaystyle (d)/(dx) = \frac{3x - 5}{2x^\bigg{(3)/(2)}}`

General Formulas and Concepts:

Algebra I

• Exponential Rule [Powering]:
  `\displaystyle (b^m)^n = b^(m \cdot n)`
• Exponential Rule [Rewrite]:
  `\displaystyle b^(-m) = (1)/(b^m)`
• Exponential Rule [Root Rewrite]:
  `\displaystyle \sqrt[n]{x} = x^{(1)/(n)}`

Calculus

Derivatives

Derivative Notation

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))`

Explanation:

Step 1: Define

Identify

`\displaystyle (3x + 5)/(√(x))`

Step 2: Differentiate

1. Rewrite [Exponential Rule - Root Rewrite]:
   `\displaystyle \frac{3x + 5}{x^\bigg{(1)/(2)}}`
2. Quotient Rule:
   `\displaystyle (d)/(dx) = \frac{(x^\bigg{(1)/(2)})(d)/(dx)[3x + 5] - (d)/(dx)[x^\bigg{(1)/(2)}](3x + 5)}{(x^\bigg{(1)/(2)})^2}`
3. Simplify [Exponential Rule - Powering]:
   `\displaystyle (d)/(dx) = \frac{(x^\bigg{(1)/(2)})(d)/(dx)[3x + 5] - (d)/(dx)[x^\bigg{(1)/(2)}](3x + 5)}{x}`
4. Basic Power Rule [Derivative Property - Addition/Subtraction]:
   `\displaystyle (d)/(dx) = \frac{(x^\bigg{(1)/(2)})(3x^(1 - 1) + 0) - ((1)/(2)x^\bigg{(1)/(2) - 1})(3x + 5)}{x}`
5. Simplify:
   `\displaystyle (d)/(dx) = \frac{3x^\bigg{(1)/(2)} - ((1)/(2)x^\bigg{(-1)/(2)})(3x + 5)}{x}`
6. Rewrite [Exponential Rule - Rewrite]:
   `\displaystyle (d)/(dx) = \frac{3x^\bigg{(1)/(2)} - (\frac{1}{2x^{(1)/(2)}})(3x + 5)}{x}`
7. Rewrite [Exponential Rule - Root Rewrite]:
   `\displaystyle (d)/(dx) = (3√(x) - ((1)/(2√(x)))(3x + 5))/(x)`
8. Simplify [Rationalize]:
   `\displaystyle (d)/(dx) = \frac{3x - 5}{2x^\bigg{(3)/(2)}}`

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

Unit: Derivatives

Book: College Calculus 10e