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

Let's suppose $y = (x)^{\frac{3}{2}}$ and let x = 4 and $\Delta x = -0.032$
Then,
$\Delta y = ({x+\Delta x})^{\frac{3}{2}} - (x)^{\frac{3}{2}}$
$\Delta y = ({4 - 0.032})^{\frac{3}{2}} - (4)^{\frac{3}{2}}$
$\Delta y = ({3.968})^{\frac{3}{2}} - 8$
$({3.968})^{\frac{3}{2}} = \Delta y + 8$
Now, we can say that $\Delta y$  is approximately equal to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{3 (x)^{\frac{1}{2}}}{2}.(-0.032) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{3}{2}} \ and \ \Delta x = -0.032)\\ dy = \frac{3 (4)^{\frac{1}{2}}}{2}.(-0.032)\\ dy = \frac{3\times 2}{2}.(-0.032)\\\\dy = -0.096$
Now,
$(3.968)^{\frac{3}{2}} = \Delta y +8\\ (3.968)^{\frac{3}{2}} = (-0.096) + 8\\ (3.968)^{\frac{3}{2}} = 7.904$
Hence, $(3.968)^{\frac{3}{2}}$ is approximately equal to 7.904

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