# Using differentials, find the approximate value of each of the following up to 3 places of decimal. $(ix)\;(82)^{\Large\frac{1}{4}}$

This is ninth 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:
Let $y=x^{\Large\frac{1}{4}}$
$y+\Delta y=(x+\Delta x)^{\Large\frac{1}{4}}$
$\Delta y=(x+\Delta x)^{\Large\frac{1}{4}}-x^{\Large\frac{1}{4}}$
$x+\Delta x=82$
$x=81=3^4$
$\Delta x=1$
Step 2:
$\Delta y=(82)^{\Large\frac{1}{4}}-(81)^{\Large\frac{1}{4}}$
$\quad\;=(82)^{\Large\frac{1}{4}}-3$
$(82)^{\Large\frac{1}{4}}=\Delta y+3$
$\Delta y\approx dy=\large\frac{dy}{dx}$$.\Delta x \Rightarrow \large\frac{1}{4}x^{-\Large\frac{3}{4}}$$\Delta x$
$\Rightarrow \large\frac{1}{4}$$(3^4)^{\Large\frac{-3}{4}}$$(1)$
$\Rightarrow \large\frac{1}{4}$$(3)^{-3}$
$\Delta y= \large\frac{1}{4\times 27}-\frac{1}{108}$
$\quad\;\;=0.009$
Step 3:
From (1) we get,
$(82)^{\Large\frac{1}{4}}=0.009+3$
$\qquad\;\;=3.009$
edited Aug 19, 2013