# Using differentials, find the approximate value of each of the following up to 3 places of decimal. $(xiii)\;(81.5)^{\Large\frac{1}{4}}$

This is thirteenth part of multipart q1

Toolbox:
• Let $y=f(x)$
• $\Delta x$ denote a small increment in $x$
• $\Delta y=f(x+\Delta x)-f(x)$
• $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
Step 1:
$y=x^{\Large\frac{1}{4}}$
$x=81$
$\Delta x=0.5$
$\Delta y=(x+\Delta x)^{\Large\frac{1}{4}}-x^{\Large\frac{1}{4}}$
$\quad\;\;=(81.5)^{\Large\frac{1}{4}}-(81)^{\Large\frac{1}{4}}$
$\quad\;\;=(81.5)^{\Large\frac{1}{4}}-3$
$(81.5)^{\Large\frac{1}{4}}=3+\Delta y$
Step 2:
$dy$ is approximately equal to $\Delta y$
$dy=\large\frac{dy}{dx}$$\Delta x \quad=\large\frac{1}{4}x^{-\Large\frac{3}{4}}$$.\Delta x$
$\quad=\large\frac{1}{4x^{\Large\frac{3}{4}}}$$.\Delta x \quad=\large\frac{1}{4(81)^{\Large\frac{3}{4}}}$$\times 0.5$
$\quad=\large\frac{0.5}{4\times 27}$
$\quad=\large\frac{0.5}{108}$
$\quad=0.0046296$
Step 3:
$(81.5)^{\Large\frac{1}{4}}=3+0.0046296$
$\qquad\;\;\;\;\;=3.0046296$
$\qquad\;\;\;\;\;=3.005$
answered Aug 5, 2013