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

- 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$

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