Question

Using differentials, find the approximate value of $\sqrt{25.2}$.

• A. $5.02$
• B. $5.01$
• C. $5.03$
• D. $5.04$

Answer 1

Answer: (A)

$\sqrt{25.2}=\sqrt{25+0.2}$
Let $x=25$ and $\Delta x=0.2$ such that $f(x)=\sqrt{x}$
$\therefore f^{\prime}(x)=\frac{1}{2 \sqrt{x}}$
$\therefore f(x+\Delta x)=f(x)+f^{\prime}(x) \cdot \Delta x$
$\sqrt{x+\Delta x}=\sqrt{x}+\frac{1}{2 \sqrt{x}} \cdot \Delta x$
$\sqrt{25+0.2}=\sqrt{25}+\frac{1}{2 \sqrt{25}} \times 0.2$
$\sqrt{25.2}=5+\frac{0.2}{2 \times 5}=5+0.02=5.02$