Question

при помощи производной вычислите приближенно 1,003^99 ; $\sqrt{4,002}$

Answer 1

1.

$f(x) = x^{99}$

$f(x) - f(x_0) \approx f'(x_0)\cdot \Delta x$

$x = 1{,}003$

$x_0 = 1$

$\Delta x = x - x_0 = 0{,}003$

$f(x) \approx f(x_0) + f'(x_0)\cdot \Delta x$

$f(x) = 1{,}003^{99}$

$f(x_0) = 1^{99} = 1$

$f'(x) = (x^{99})' = 99\cdot x^{98}$

$f'(x_0) = 99\cdot x_0^{98} = 99\cdot 1^{98} = 99$

$1{,}003^{99} \approx 1 + 99\cdot 0{,}003 = 1 + 0{,}297 = 1{,}297$

2.

$f(x) = \sqrt{x}$

$f(x) - f(x_0) \approx f'(x_0)\cdot \Delta x$

$x = 4{,}002$

$x_0 = 4$

$\Delta x = x - x_0 = 0{,}002$

$f'(x) = (\sqrt{x})' = \frac{1}{2\cdot\sqrt{x}}$

$f'(x_0) = \frac{1}{2\cdot\sqrt{x_0}} = \frac{1}{2\cdot\sqrt{4}} = \frac{1}{2\cdot 2} =$

$= \frac{1}{4} = 0{,}25$

$\sqrt{4{,}002} \approx \sqrt{4} + 0{,}25\cdot 0{,}002 =$

$= 2 + 0{,}0005 = 2{,}0005$