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

Question

This is ninth part of multipart q1

Answer 1

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,

$(82)^{\Large\frac{1}{4}}=0.009+3$

$\qquad\;\;=3.009$