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# Using differentials, find the approximate value of each of the following up to 3 places of decimal. $(xi)\;(0.0037)^{\Large\frac{1}{2}}$

This is eleventh part of multipart q1

Can you answer this question?

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 $y=\sqrt x$
Also Let $x=0.0036$ and $\Delta x=0.0001$
So that $x+\Delta x=0.0037$
$\Delta y=\sqrt{x+\Delta x}-\sqrt x$
$\quad=\sqrt{0.0037}-\sqrt{0.0036}$
$\quad=\sqrt{0.0037}-0.06$
$\sqrt{0.0037}=0.06+\Delta y$
Also $\large\frac{dy}{dx}=\frac{1}{2}x^{\Large\frac{1}{2}}$
$\qquad\;\;\;\;=\large\frac{1}{2\sqrt x}$
Step 2:
$\Delta y$ is approximately equal to $dy$
$dy=\big(\large\frac{dy}{dx}\big)$$\Delta x \quad=\large\frac{1}{2\sqrt x}$$\Delta x$
$\quad=\large\frac{1}{2\sqrt{0.0036}}$$\times 0.0001$
$\quad=\large\frac{0.0001}{2\times 0.06}$
$\quad=\large\frac{0.0001}{0.12}$
$\quad=\large\frac{0.01}{12}$
$\quad=0.000833$
Step 3:
Approximate value of $\Delta y=dy=0.000833$
Hence from (1) we have
$\sqrt{0.0037}=0.06+0.00083$
$\qquad\;\;\;\;\;\;=0.061$
answered Aug 5, 2013