# FInd the approximate value of $(1.999)^5$.

$\begin{array}{1 1} 31 \\ 31.92 \\ 30 \\ 31.82 \end{array}$

Toolbox:
• If $y = f(x)$, then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
• $\bigg(\large \frac{dy}{dx} \bigg)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1
$(1.999)^5$
Let $y=f(r)=(1.999)^5$
Let $x=2\: and \: x-\Delta x=1.999$
$\therefore \Delta x=1.999+0.001=2$
$\therefore x=2\: and\: \Delta x=0.001$
For $x=2$, we have $y=(2)^5$
$=32$
$dx=\Delta x=-0.001$
Step 2
$y=(x)^5$
Differentiating on both sides w.r.t $x$,
$\large\frac{dy}{dx}=5x^4$
$\Rightarrow dy=5x^4.dx$
When $x=2\: and \: dx=-0.001$
$dy=5(2)^4.(-0.001)$
$= -80 \times 0.001$
$= -.08 \Rightarrow \Delta y=-0.08$
Hence $(1.999)^5=y-\Delta y$
=$32-0.08$
$= 31.92$