Упр.35.8 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) f(x)=xvx, x_0=81; 3) f(x)=v(xvx), x_0=16;
2) f(x)=x^3 x^(1/4), x_0=1; 4) f(x)=x^2/x^(1/6), x_0=64.

1. $$f(x)=x\sqrt{x}=x^{\frac32}$$

   $$f'(x)=\left(x^{\frac32}\right)’=\frac32x^{\frac12}=\frac{3\sqrt{x}}{2}$$

   $$f'(81)=\frac{3\sqrt{81}}{2}=\frac{3\cdot 9}{2}=\frac{27}{2}=13{,}5$$

2. $$f(x)=x^3\sqrt[4]{x}=x^3\cdot x^{\frac14}=x^{\frac{13}{4}}$$

   $$f'(x)=\left(x^{\frac{13}{4}}\right)’=\frac{13}{4}x^{\frac94}$$

   $$f'(1)=\frac{13}{4}\cdot 1^{\frac94}=\frac{13}{4}=3{,}25$$

3. $$f(x)=\sqrt{x\sqrt{x}}=\sqrt{x\cdot x^{\frac12}}=\sqrt{x^{\frac32}}=x^{\frac34}$$

   $$f'(x)=\left(x^{\frac34}\right)’=\frac34x^{-\frac14}=\frac{3}{4\sqrt[4]{x}}$$

   $$f'(16)=\frac{3}{4\sqrt[4]{16}}=\frac{3}{4\cdot 2}=\frac38$$

4. $$f(x)=\frac{x^2}{\sqrt[6]{x}}=x^2\cdot x^{-\frac16}=x^{\frac{11}{6}}$$

   $$f'(x)=\left(x^{\frac{11}{6}}\right)’=\frac{11}{6}x^{\frac56}$$

   $$f'(64)=\frac{11}{6}\cdot 64^{\frac56}=\frac{11}{6}\cdot (2^6)^{\frac56}=\frac{11}{6}\cdot 2^5=\frac{11\cdot 32}{6}=\frac{176}{3}$$

Ответ

1) $$13{,}5$$; 2) $$3{,}25$$; 3) $$\frac38$$; 4) $$\frac{176}{3}$$.