1.Найдите производные функций: а) f(x)=5x^4+3x^2-8x-9 б) g(x)=1/x*√x в) q(x)=3x-2 -------...

№ 1
а) $f(x)=5x^4+3x^2-8x-9$
$f'(x)=(5x^4+3x^2-8x-9)'=4*5x^3+2*3x-8=20x^3+6x-8$

б) $g(x)= \frac{1}{x} * \sqrt{x}$
$\frac{1}{x} * \sqrt{x} = \frac{ \sqrt{x} }{x}= \frac{1}{ \sqrt{x} }= \frac{1}{x^{0.5}} =x^{-0.5}$
$g(x)=x^{-0.5}$
$g'(x)=(x^{-0.5})'=-0.5x^{-1.5}=- \frac{1}{2\sqrt{x^3} }$

в) $q(x)= \frac{3x-2}{x+3}$
$q'(x)= (\frac{3x-2}{x+3} )'= \frac{(3x-2)'*(x+3)-(3x-2)*(x+3)'}{(x+3)^2} = \frac{3(x+3)-1*(3x-2)}{(x+3)^2} =$ $=\frac{3x+9-3x+2}{(x+3)^2}= \frac{11}{(x+3)^2}$

г) $u(x)=sin 5x$
$u'(x)=(sin 5x)'=cos5x*(5x)'=5cos5x$

№ 2
$f(x)= \frac{2x-1}{x}$ $x_0=2$
$f'(x)= (\frac{2x-1}{x} )'= \frac{(2x-1)'*x-(2x-1)*x'}{x^2} = \frac{2x-(2x-1)}{x^2} = \frac{2x-2x+1}{x^2}= \frac{1}{x^2}$
$f'(2)= \frac{1}{2^2} = \frac{1}{4}$
$tg \alpha =f'(x_0)$
$tg \alpha = \frac{1}{4}$
$\alpha =arctg \frac{1}{4}$