Question

Using differentiation find the approximate value of
$\sqrt{0.037}$
​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm :\longmapsto\:f(x) = \sqrt{x} \: \: where \: x = 0.0361$

So,

$\rm :\longmapsto\:f(x + h) = \sqrt{x + h} \: \: where \: x = 0.0009$

By using definition, we get

$\rm :\longmapsto\:\boxed{ \tt{ \: f(x + h) = f(x) + h \times f'(x) \: }}$

So, on substituting the values, we get

$\rm :\longmapsto\: \sqrt{x + h} = \sqrt{x} + h \times \dfrac{d}{dx} \sqrt{x}$

$\rm :\longmapsto\: \sqrt{x + h} = \sqrt{x} + h \times \dfrac{1}{2 \sqrt{x} }$

On substituting the values of x and h, we get

$\rm :\longmapsto\: \sqrt{0.0361 + 0.0009} = \sqrt{0.0361} + \dfrac{0.0009}{2 \times \sqrt{0.0361} }$

$\rm :\longmapsto\: \sqrt{0.037} = \sqrt{\dfrac{361}{10000} } + \dfrac{0.0009}{2 \times \sqrt{\dfrac{361}{10000} } }$

$\rm :\longmapsto\: \sqrt{0.037} = \dfrac{19}{100} + \dfrac{9}{10000 \times 2 \times \dfrac{19}{100} }$

$\rm :\longmapsto\: \sqrt{0.037} = 0.19+ \dfrac{9}{3800}$

$\rm :\longmapsto\: \sqrt{0.037} = 0.19+ 0.0024$

$\rm :\longmapsto\: \sqrt{0.037} = 0.1924$

$\pink{\rm \implies\:\boxed{ \tt{ \: \: \sqrt{0.037} = 0.1924}}}$

More to know :-

$\pink{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}}$