Question

Using the differentials, the approximate value of $(627{)}^{1/4}$ is

• A. 5.002
• B. 5.003
• C. 5.005
• D. 5.004

Answer 1

Answer: (D)

5.004

$(627{)}^{1/4}=(625+2{)}^{1/4}$
Let $y=f\left(x\right)=x^{1/4}$
$y+\Delta y=(x+\Delta x{)}^{1/4}$
$\Rightarrow \Delta y=(x+\Delta x{)}^{1/4}-x^{1/4}$
Also, $\Delta y=f^{\prime }\left(x\right)\Delta x$
$\Rightarrow (x+\Delta x{)}^{1/4}-x^{1/4}=\frac{1}{4}{x}^{-3/4}\Delta x$
Put $x=625,\Delta x=2$
$\Rightarrow (627{)}^{1/4}-5=\frac{1}{4}(5{)}^{-3}\left(2\right)$
$\Rightarrow (627{)}^{1/4}=5+0.004=5.004$