Question

Differentiate Vx(x + 2) with respect to x.

Answer 1

Answer:

$y'= \frac{3x+2}{2\sqrt{x} }$

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: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

$y=\sqrt{x} (x+2)$

Step 2: Rewrite

$y=x^{\frac{1}{2} } (x+2)$

Step 3: Differentiate

1. Product Rule [Basic Power/Chain Rule]:                 $y'= \frac{1}{2} x^{\frac{1}{2} -1}(x+2)+x^{\frac{1}{2} }(1)$
2. Simplify:                                                                            $y'= \frac{1}{2} x^{\frac{-1}{2}}(x+2)+x^{\frac{1}{2}}$
3. Rewrite:                                                                                        $y'= \frac{x+2}{2\sqrt{x} } + \sqrt{x}$
4. Add:                                                                                                       $y'= \frac{3x+2}{2\sqrt{x} }$