Question

Derivative, how to get the derivative of a root? example: √5x-6

Answer 1

To find the derivative of a square root, you will just need to consider the expression under the square root as a whole term, or in calculus, as a variable called 'u.'

In this case, our expression is: √(5x-6)

Let u = 5x - 6

du = 5

d/du (√u) = du * 0.5 * u^(-1/2)

Substitute back for u:

d/dx (√u) = 5 * 0.5 * (5x - 6)^(-1/2)

d/dx (√u) = 2.5/√(5x - 6)

Hope this helped!

Answer 2

`\displaystyle (d)/(dx)[√(5x - 6)] = (5)/(2√(5x - 6))`

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

Explanation:

Step 1: Define

Identify

`\displaystyle y = √(5x - 6)`

Step 2: Differentiate

1. [Function] Rewrite:
   `\displaystyle y = (5x - 6)^\Big{(1)/(2)}`
2. [Function] Basic Power Rule [Derivative Rule - Chain Rule]:
   `\displaystyle y' = (1)/(2)(5x - 6)^\Big{-(1)/(2)} (d)/(dx)[5x - 6]`
3. Rewrite:
   `\displaystyle y' = \frac{1}{2(5x - 6)^\Big{(1)/(2)}}(d)/(dx)[5x - 6]`
4. Rewrite [Derivative Rule - Addition/Subtraction]:
   `\displaystyle y' = \frac{1}{2(5x - 6)^\Big{(1)/(2)}} \bigg( (d)/(dx)[5x] - (d)/(dx)[6] \bigg)`
5. Rewrite [Derivative Rule - Multiplied Constant]:
   `\displaystyle y' = \frac{1}{2(5x - 6)^\Big{(1)/(2)}} \bigg( 5 (d)/(dx)[x] - (d)/(dx)[6] \bigg)`
6. Basic Power Rule:
   `\displaystyle y' = \frac{5}{2(5x - 6)^\Big{(1)/(2)}}`
7. Rewrite:
   `\displaystyle y' = (5)/(2√(5x - 6))`

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

Unit: Differentiation