# Using differentials, find the approximate value of each of the following up to $3$ places of decimal. $(255)^{\Large\frac{1}{4}}$

$\begin{array}{1 1} 3.9963 \\ 3.9961 \\ 4.9963 \\ 2.9963 \end{array}$

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 $y=f(x)=x^{\Large\frac{1}{4}}$
Let $x=256$
$x+\Delta x=255$
So that $\Delta x=255-256=-1$
$\Delta y=(x+\Delta x)^{\Large\frac{1}{4}}-x^{\Large\frac{1}{4}}$
$\Delta y=(256-1)^{\Large\frac{1}{4}}-(256)^{\Large\frac{1}{4}}$
$\quad=(255)^{\Large\frac{1}{4}}-4$
$(256)^{\Large\frac{1}{4}}=4+\Delta y$
Step 2:
Also $\large\frac{dy}{dx}=\frac{1}{4}x^{\Large\frac{1}{4}-1}$
$\qquad\;\;=\large\frac{1}{4}x^{-\Large\frac{3}{4}}$
$\qquad\;\;=\large\frac{1}{4x^{\Large\frac{3}{4}}}$
Now $\Delta y$ is approximately equal to $dy$
$dy=\big(\large\frac{dy}{dx}\big)$$dx \quad=\large\frac{1}{4x^{\Large\frac{3}{4}}}$$dx$
$\quad=\large\frac{1}{4(256)^{\Large\frac{3}{4}}}$$(-1) \quad=\large\frac{-1}{4(4^4)^{\Large\frac{3}{4}}}$$dx$
$\quad=\large\frac{1}{4\times 64}$
$\Rightarrow \large\frac{-1}{256}=$$-0.0039$
Step 3:
Approximate value of $\Delta y=dy=-0.0039$
Hence from (1) we have
$(255)^{\Large\frac{1}{4}}=4-0.0039$
$\qquad\quad\;=3.9961$
answered Aug 5, 2013