Question

$f(x) = (4x + 3) ^{6} \ x0 =-1$
$f(x) = 2 - 2 cos(x) \ x0 = frac{pi}{6}$
$f(x) = sqrt{x^{2} - 8 } \ x0 = 3$
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Answer 1

1)

$f'(x) = 24(4x+3)^5$

$f'(x_0) = 24(-4+3)^5=-24$

2)

$f'(x)=-2(-sin x)=2sin x$

$f'(x_0)=2sin frac{pi}{6} =1$

3)

$f'(x) = frac{1}{2sqrt{x^2-8}}2x = frac{x}{sqrt{x^2-8}}$

$f'(x_0) = frac{3}{sqrt{3^2-8}}=3$

4)

$f'(x)=frac{1}{2}2cos 2x = cos 2x$

$f'(x_0) = cos 2frac{pi}{8} cos frac{pi}{4} =frac{sqrt{2}}{2}$

5)

$f'(x) = 9(3x-5)^2-frac{2(-1)}{(3-x)^3} = 9(3x-5)^2+frac{2}{(3-x)^3}$

$f'(x_0) = 9(6-5)^2+frac{2}{(3-2)^3} =9 +2=11$