Question

Помогите с математикой пожалуйста 11 класс
*на фото (№69: 1 и 2)

Answer 1

$1)\ \ f(x)=\dfrac{x^3}{\sqrt{8+x^3}}\\\\\\f'(x)=\dfrac{3x^2\cdot \sqrt{8+x^3}-x^3\cdot \frac{1}{2\sqrt{8+x^3}}\cdot 3x^2}{8+x^3}=\dfrac{6x^2\, (8+x^3)-3x^5}{2\sqrt{(8+x^3)^3}}=\dfrac{48x^2+3x^5}{2\sqrt{(8+x^3)^3}}\\\\\\f'(1)=\dfrac{48+3}{2\sqrt{9^3}}=\dfrac{51}{2\cdot 9\cdot 3}=\dfrac{51}{54}$

$2)\ \ f(x)=\sqrt{\dfrac{\sqrt{x}-1}{\sqrt{x}+1}}\\\\\\f'(x)=\dfrac{1}{2\sqrt{\frac{\sqrt{x}-1}{\sqrt{x}+1}}}\cdot \dfrac{\frac{1}{2\sqrt{x}}\cdot (\sqrt{x}+1)-\frac{1}{2\sqrt{x}}\cdot (\sqrt{x}-1)}{(\sqrt{x}+1)^2}=\\\\\\=\dfrac{\sqrt{\sqrt{x}+1}}{2\sqrt{\sqrt{x}-1}}\cdot \dfrac{\sqrt{x}+1-\sqrt{x}+1}{2\sqrt{x}\cdot (\sqrt{x}+1)^2}=\dfrac{\sqrt{\sqrt{x}+1}}{2\sqrt{\sqrt{x}-1}}\cdot \dfrac{2}{2\sqrt{x}\cdot (\sqrt{x}+1)^2}=$

$=\dfrac{\sqrt{\sqrt{x}+1}}{2\sqrt{\sqrt{x}-1}\cdot \sqrt{x}\cdot (\sqrt{x}+1)^2}=\dfrac{1}{2\sqrt{\sqrt{x}-1}\cdot \sqrt{x}\cdot \sqrt{(\sqrt{x}+1)^3}}=\\\\\\=\dfrac{1}{2\sqrt{(\sqrt{x}-1)(\sqrt{x}+1)}\cdot \sqrt{x}\cdot (\sqrt{x}+1)}=\dfrac{1}{2\sqrt{x}\cdot \sqrt{x-1}\cdot (\sqrt{x}+1)}$