Страница 243 номер 811, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(f(x) = (x - 1)^{8}(2 - x)^{7}\)
\(f'(x) = ((x - 1)^{8})'(2 - x)^{7} + (x - 1)^{8}((2 - x)^{7})' =\)
\(= 8(x - 1)^{7}(2 - x)^{7} - 7(x - 1)^{8}(2 - x)^{6},\)
\(f'(1) = 8(1 - 1)^{7}(2 - 1)^{7} - 7(1 - 1)^{8}(2 - 1)^{6} = 0;\)

2) \(f(x) = (2x - 1)^{5}(1 + x)^{4}\)
\(f'(x) = ((2x - 1)^{5})'(1 + x)^{4} + (2x - 1)^{5}((1 + x)^{4})' =\)
\(= 5 \cdot 2(2x - 1)^{4}(1 + x)^{4} + 4(2x - 1)^{5}(1 + x)^{3},\)
\(f'(1) = 10(2 - 1)^{4}(1 + 1)^{4} + 4(2 - 1)^{5}(1 + 1)^{3} =\)
\(= 10 \cdot 16 + 4 \cdot 8 = 192;\)

3) \(f(x) = \sqrt{2 - x}(3 - 2x)^{8}\)
\(f'(x) = \left(\sqrt{2 - x}\right)'(3 - 2x)^{8} + \sqrt{2 - x}((3 - 2x)^{8})' =\)
\(= -\dfrac{(3 - 2x)^{8}}{2\sqrt{2 - x}} - 2 \cdot 8 \cdot \sqrt{2 - x}(3 - 2x)^{7},\)
\(f'(1) = -\dfrac{(3 - 2)^{8}}{2} - 16 \cdot (3 - 2)^{7} = -\dfrac{1}{2} - 16 = -16\dfrac{1}{2};\)

4) \(f(x) = (5x - 4)^{6}\sqrt{3x - 2}\)
\(f'(x) = ((5x - 4)^{6})'\sqrt{3x - 2} + (5x - 4)^{6}\left(\sqrt{3x - 2}\right)' =\)
\(= 6 \cdot 5 \cdot (5x - 4)^{5}\sqrt{3x - 2} + \dfrac{3(5x - 4)^{6}}{2\sqrt{3x - 2}},\)
\(f'(1) = 30 \cdot (5 - 4)^{5}\sqrt{3 - 2} + \dfrac{3(5 - 4)^{6}}{2\sqrt{3 - 2}} = 30 + \dfrac{3}{2} = \dfrac{63}{2}.\)