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# Using differentials, find the approximate value of $\sqrt {49.5}$.  Verified
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Hint: The differentiation of $x$ is represented by $dx$ is defined by $dx = x$ where $x$ is the minor change in $x$. The differential of $y$ is represented by $dy$ is defined by $dy = \dfrac{{dy}}{{dx}}x$. As $x$ is very small compared to $x$, so $dy$ is the approximation of $y$. Hence the increment in $y$ corresponding to the increment in $x$, denoted by $\Delta y$, is given by $\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right)$.

Let $y = \sqrt x$ where $x = 49\& \Delta x = 0.5$
Since $y = \sqrt x$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\sqrt x } \right)}}{{dx}} = \dfrac{1}{{2\sqrt x }}$
$\Delta y = \dfrac{{dy}}{{dx}}\Delta x \\ \Rightarrow \Delta y = \dfrac{1}{{2\sqrt x }}\left( {0.5} \right) \\ \Rightarrow \Delta y = \dfrac{1}{{2\sqrt {49} }} \times 0.5 \\ \Rightarrow \Delta y = \dfrac{1}{{2 \times 7}} \times 0.5 \\ \Rightarrow \Delta y = \dfrac{1}{{14}} \times 0.5 \\ \Rightarrow \Delta y = \dfrac{{0.5}}{{14}} \\ \therefore \Delta y = 0.036 \\$
$\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right)$
$\Delta y = \sqrt {x + \Delta x} - \sqrt x {\text{ }}\left[ {\because f\left( x \right) = y = \sqrt x } \right]{\text{ }} \\ 0.036 = \sqrt {49 + 0.5} - \sqrt {49} \\ {\text{0}}{\text{.036}} = \sqrt {49.5} - 7 \\ \sqrt {49.5} = 0.036 + 7 \\ \therefore \sqrt {49.5} = 7.036{\text{ }} \\$
Thus, the approximate value of $\sqrt {49.5}$ is 7.036