Question

.. Срочно 100 б.......................​

Answer 1

$\displaystyle\bf\\f(x)=(x^{2} +3x)\sqrt{x} \\\\f'(x)=(x^{2} +3x)'\cdot \sqrt{x} +(x^{2} +3x)\cdot (\sqrt{x} )'=(2x+3)\cdot \sqrt{x} +(x^{2} +3x)\cdot\frac{1}{2\sqrt{x} } )\\\\\\=(2x+3)\cdot \sqrt{x} +\frac{x(x+3)}{2\sqrt{x} } =(2x+3)\cdot \sqrt{x} +0,5\sqrt{x} \cdot(x+3)\\\\\\f'(9)=(2\cdot 9+3)\cdot\sqrt{9} +0,5\sqrt{9} \cdot(9+3) =21\cdot 3+0,5\cdot 3\cdot 12=\\\\\\=63+18=81\\\\\\\boxed{f'(9)=81}$

$\displaystyle\bf\\f'\Big(\frac{1}{4} \Big)=\Big(2\cdot\frac{1}{4} +3\Big)\cdot\sqrt{\frac{1}{4} } +0,5\cdot\sqrt{\frac{1}{4} } \cdot \Big(\frac{1}{4} +3\Big)=\\\\\\=3,5\cdot 0,5+0,5\cdot 0,5 \cdot 3,25=1,75+0,8125=2,5625\\\\\\f'\Big(\frac{1}{4} \Big)=2,5625$

Answer 2

Ответ:

$f(x)=(x^2+3x)\sqrt{x}$

Производная произведения:  $(uv)'=u'v+uv'$  .

$f'(x)=(2x+3)\sqrt{x}+(x^2+3x)\cdot \dfrac{1}{2\sqrt{x}}\\\\\\a)\ \ f'(9)=(18+3)\cdot 3+\dfrac{81+27}{2\cdot 3}=21\cdot 3+18=81\\\\b)\ \ f'(\dfrac{1}{4})=(\dfrac{1}{2}+3)\cdot \dfrac{1}{2}+(\dfrac{1}{16}+\dfrac{3}{4})\cdot \dfrac{1}{2\cdot \frac{1}{2}}=\dfrac{7}{4}+\dfrac{13}{16}=\dfrac{41}{16}=2,5625$