Упр.230 ГДЗ Колмогоров 10-11 класс (Алгебра)

Найдём производные по правилу производной сложной функции.

1. $$f(x)=(x^3-2x^2+3)^{17}$$
   $$f'(x)=17(x^3-2x^2+3)^{16}\cdot(3x^2-4x)$$
2. $$f(x)=\sqrt{1-x^4}+\frac{1}{x^2+3}$$
   $$f'(x)=\frac{1}{2}(1-x^4)^{-\frac12}\cdot(-4x^3)-\frac{2x}{(x^2+3)^2}$$
   $$f'(x)=-\frac{2x^3}{\sqrt{1-x^4}}-\frac{2x}{(x^2+3)^2}$$
3. $$f(x)=\sqrt{4x^2+5}$$
   $$f'(x)=\frac{1}{2}(4x^2+5)^{-\frac12}\cdot 8x=\frac{4x}{\sqrt{4x^2+5}}$$
4. $$f(x)=(3-x^3)^5+\sqrt{2x-7}$$
   $$f'(x)=5(3-x^3)^4\cdot(-3x^2)+\frac{1}{2}(2x-7)^{-\frac12}\cdot 2$$
   $$f'(x)=-15x^2(3-x^3)^4+\frac{1}{\sqrt{2x-7}}$$

Ответ:

$$f'(x)=17(x^3-2x^2+3)^{16}(3x^2-4x),$$
$$f'(x)=-\frac{2x^3}{\sqrt{1-x^4}}-\frac{2x}{(x^2+3)^2},$$
$$f'(x)=\frac{4x}{\sqrt{4x^2+5}},$$
$$f'(x)=-15x^2(3-x^3)^4+\frac{1}{\sqrt{2x-7}}.$$