Respuesta :
Answer:
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:
[tex]\displaystyle \frac{d}{dx}[\sqrt{5x - 6}] = \frac{5}{2\sqrt{5x - 6}}[/tex]
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle y = \sqrt{5x - 6}[/tex]
Step 2: Differentiate
- [Function] Rewrite: [tex]\displaystyle y = (5x - 6)^\Big{\frac{1}{2}}[/tex]
- [Function] Basic Power Rule [Derivative Rule - Chain Rule]: [tex]\displaystyle y' = \frac{1}{2}(5x - 6)^\Big{-\frac{1}{2}} \frac{d}{dx}[5x - 6][/tex]
- Rewrite: [tex]\displaystyle y' = \frac{1}{2(5x - 6)^\Big{\frac{1}{2}}}\frac{d}{dx}[5x - 6][/tex]
- Rewrite [Derivative Rule - Addition/Subtraction]: [tex]\displaystyle y' = \frac{1}{2(5x - 6)^\Big{\frac{1}{2}}} \bigg( \frac{d}{dx}[5x] - \frac{d}{dx}[6] \bigg)[/tex]
- Rewrite [Derivative Rule - Multiplied Constant]: [tex]\displaystyle y' = \frac{1}{2(5x - 6)^\Big{\frac{1}{2}}} \bigg( 5 \frac{d}{dx}[x] - \frac{d}{dx}[6] \bigg)[/tex]
- Basic Power Rule: [tex]\displaystyle y' = \frac{5}{2(5x - 6)^\Big{\frac{1}{2}}}[/tex]
- Rewrite: [tex]\displaystyle y' = \frac{5}{2\sqrt{5x - 6}}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation