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# Using differentials, find the approximate value of each of the following up to $3$ places of decimal. $(3.968)^{\Large\frac{3}{2}}$

$\begin{array}{1 1} 2.904 \\ 7.904 \\ 4.904 \\ 5.904 \end{array}$

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A)
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 $y=x^{\Large\frac{3}{2}}$
$x=4$
$\Delta x=-0.032$
$\Delta y=(x+\Delta x)^{\Large\frac{3}{2}}-x^{\Large\frac{3}{2}}$
$\quad\;\;=(4-0.032)^{\Large\frac{3}{2}}-4^{\Large\frac{3}{2}}$
$\quad\;\;=(3.968)^{\Large\frac{3}{2}}-8$
$(3.968)^{\Large\frac{3}{2}}$$=8+\Delta y------(1) Step 2: dy is the approximate value of \Delta y dy=\large\frac{dy}{dx}$$\times \Delta x$
$\quad=\large\frac{3}{2}x^{\Large\frac{1}{2}}.\Delta x$
$\quad=\large\frac{3}{2}$$\sqrt x.\Delta x \Rightarrow \large\frac{3}{2}$$\sqrt 4\times (-0.032)$
$\Rightarrow \large\frac{3}{2}$$2\times (-0.032)$
$\Rightarrow 3\times (-0.032)$
$\Rightarrow -0.096)$
Step 3:
From (1) we have
$(3.968)^{\Large\frac{3}{2}}=8+\Delta y$
$\qquad\;\;\;\;\;\;\;=8-0.096$
$\qquad\;\;\;\;\;\;\;=7.904$