If $f\left(x\right)=\sqrt{x+2}$, find the differential $dy$ when $x=2$ and $dx=0.01$.

• A. 0.02
• B. 0.0025
• C. 2.00
• D. 0.025

Answer: (B)

Verified step by step guidance

1. First, identify the function given: \( f(x) = \sqrt{x + 2} \).
2. To find the differential \( dy \), we need to compute the derivative \( f'(x) \). Use the chain rule to differentiate \( f(x) = \sqrt{x + 2} \). The derivative is \( f'(x) = \frac{1}{2\sqrt{x + 2}} \).
3. Evaluate the derivative at \( x = 2 \). Substitute \( x = 2 \) into \( f'(x) \) to get \( f'(2) = \frac{1}{2\sqrt{2 + 2}} = \frac{1}{2\sqrt{4}} = \frac{1}{4} \).
4. The differential \( dy \) is given by \( dy = f'(x) \cdot dx \). Substitute \( f'(2) = \frac{1}{4} \) and \( dx = 0.01 \) into this formula to get \( dy = \frac{1}{4} \times 0.01 \).
5. Simplify the expression for \( dy \) to find the approximate change in \( y \) when \( x \) changes by \( dx = 0.01 \).