Question

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

Answer 1

1)

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

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

2)

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

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

3)

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

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

4)

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

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

5)

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

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