Question

Using differentiation find the approximate value of

${(1.999)}^{5}$
​

Answer 1

Answer:

the correct answer is 31.9 that is 32 approximately

Answer 2

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

Let assume that

$\rm :\longmapsto\:f(x) = {x}^{5} \: \: \: where \: x = 2$

So,

$\rm :\longmapsto\:f(x + h) = {(x + h)}^{5} \: \: \: where \: x = - 0.001$

By using definition of differentiation, we have

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

On substituting the values, we get

$\rm :\longmapsto\: {(x + h)}^{5} = {x}^{5} + h \times \dfrac{d}{dx} {x}^{5}$

$\rm :\longmapsto\: {(x + h)}^{5} = {x}^{5} + h \times {5x}^{5 - 1}$

$\rm :\longmapsto\: {(x + h)}^{5} = {x}^{5} + h \times {5x}^{4}$

$\rm :\longmapsto\: {(x + h)}^{5} = {x}^{5} + {5hx}^{4}$

On substituting the values of x and h, we get

$\rm :\longmapsto\: {(2 - 0.001)}^{5} = {2}^{5} + (5)( - 0.001) {2}^{4}$

$\rm :\longmapsto\: {(1.999)}^{5} = 32 - 0.08$

$\rm :\longmapsto\: {(1.999)}^{5} = 31.92$

Hence,

$\purple{\rm \implies\:\: \boxed{ \tt{ \: {(1.999)}^{5} = 31.92 \: }}}$

More to know :-

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