Question

Найдите производные функции f(x), если
a) f(x)=4x^7-13
б) f(x)=14x^3+5x^-12-2x^8
в) f(x)=(9x-4)*(8x^2+3)
г) f(x)=-2 cos(7x-1)+9x^2
д) y=4√25-3x

Answer 1

$\displaystyle\bf\\1)\\\\f(x)=4x^{7} -13\\\\f'(x)=4\cdot(x^{7})'-13'=4\cdot 7x^{6}-0=28x^{6} \\\\\\2)\\\\f(x)=14x^{3}+5x^{-12} -2x^{8}\\\\f'(x)=14\cdot(x^{3} )'+5\cdot(x^{-12} )'-2\cdot(x^{8} )'=14\cdot 3x^{2} +5\cdot(-12x^{-13} )-2\cdot 8x^{7} =\\\\=42x^{2} -60x^{-13} -16x^{7} \\\\\\3)\\\\f(x)=(9x-4)\cdot(8x^{2} +3)\\\\f'(x)=(9x-4)'\cdot(8x^{2} +3)+(9x-4)\cdot(8x^{2} +3)'=9\cdot(8x^{2} +3)+(9x-4)\cdot 16x=\\\\=72x^{2} +27+144x^{2} -64x=216x^{2} -64x+27$

$\displaystyle\bf\\4)\\\\f(x)=-2Cos(7x-1)+9x^{2} \\\\f'(x)=-2\cdot\Big[Cos(7x-1)\Big]'+9\cdot(x^{2} )'=-2\cdot\Big[-Sin(7x-1)\Big]\cdot(7x-1)'+9\cdot 2x=\\\\=2Sin(7x-1)\cdot 7+18x=14Sin(7x-1)+18x\\\\\\5)\\\\y=4\sqrt{25-3x} \\\\y'=4\cdot(\sqrt{25-3x} )'=4\cdot\frac{1}{2\sqrt{25-3x} } \cdot(25-3x)'=\frac{2}{\sqrt{25-3x} } \cdot(-3)=\\\\=-\frac{6}{\sqrt{25-3x} }$