Найти f `(x0) , если f(x) = 1/2 sin2x, x0= п/8 Найти f `(x0) , если f(x) = (3x-5)^3 + 1/ (3-x)^2, x0=2. Created by wwwivanfedin9. algebra-ru

(3-x)^2, x0=2...

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Найти f `(x0) , если f(x) = 1/2 sin2x, x0= п/8

Найти f `(x0) , если f(x) = (3x-5)^3 + 1/ (3-x)^2, x0=2

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Luluput

Verified answer

1)
$f(x) = \frac{1}{2} sin2x,$ $x_0= \frac{ \pi }{8}$
$f'(x) =( \frac{1}{2} sin2x)'= \frac{1}{2} cos2x*(2x)'= \frac{1}{2} cos2x*2=cos2x$
$f'( \frac{ \pi }{8} )=cos(2* \frac{ \pi }{8}) =cos \frac{ \pi }{4} = \frac{ \sqrt{2} }{2}$

2)
$f(x) = (3x-5)^3 + \frac{1}{(3-x)^2},$ $x_0=2$
$f'(x) =( (3x-5)^3 + \frac{1}{(3-x)^2} )'=((3x-5)^3)' + ( \frac{1}{(3-x)^2} )'=$ $=3*(3x-5)^2*3+ \frac{0-2(3-x)*(-1)}{(3-x)^4} =9*(3x-5)^2+ \frac{2(3-x)}{(3-x)^4} =$ $=9*(3x-5)^2+ \frac{2}{(3-x)^3}$
$f'(2)=9*(3*2-5)^2+ \frac{2}{(3-2)^3}=9+2=11$

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