Answer:
[tex]y'= \frac{3x+2}{2\sqrt{x} }[/tex]
General Formulas and Concepts:
Calculus
The derivative of a constant is equal to 0
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Product Rule: [tex]\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)[/tex]
Chain Rule: [tex]\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Step-by-step explanation:
Step 1: Define
[tex]y=\sqrt{x} (x+2)[/tex]
Step 2: Rewrite
[tex]y=x^{\frac{1}{2} } (x+2)[/tex]
Step 3: Differentiate
- Product Rule [Basic Power/Chain Rule]: [tex]y'= \frac{1}{2} x^{\frac{1}{2} -1}(x+2)+x^{\frac{1}{2} }(1)[/tex]
- Simplify: [tex]y'= \frac{1}{2} x^{\frac{-1}{2}}(x+2)+x^{\frac{1}{2}}[/tex]
- Rewrite: [tex]y'= \frac{x+2}{2\sqrt{x} } + \sqrt{x}[/tex]
- Add: [tex]y'= \frac{3x+2}{2\sqrt{x} }[/tex]
