Question

Find the approximation of √97 using derivatives.

This is of application for derivative​

Answer 1

97 is not a perfect square and hence its square root is not a whole number. √97 = 9.848857 (approx)

ANSWER BY OM

Answer 2

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

Let assume that,

$\rm :\longmapsto\:f(x) = \sqrt{x} \: \: \: \: \: \: where \: x = 100$

So,

$\rm :\longmapsto\:f(x + h) = \sqrt{x + h} \: \: \: \: \: \: where \: h = - \: 3$

By using definition of differentiation, we have

$\rm :\longmapsto\:f(x + h) = f(x) + h \: f'(x)$

On substituting the values, we get

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

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

On substituting x = 100, h = - 3

$\rm :\longmapsto\: \sqrt{100 - 3} = \sqrt{100} - 3 \times \dfrac{1}{2 \sqrt{100} }$

$\rm :\longmapsto\: \sqrt{97} = 10 - \dfrac{3}{20}$

$\rm :\longmapsto\: \sqrt{97} = \dfrac{200 - 3}{20}$

$\rm :\longmapsto\: \sqrt{97} = \dfrac{197}{20}$

$\rm :\longmapsto\: \sqrt{97} = 9.85$

$\bf\implies \:\boxed{ \tt{ \: \: \sqrt{97} \: = \: 9.85 \: \: (approx.) \: \: }}$

More to know :-

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