Question

$1)f(x)=sin^{5}x, x_0= \frac{ \pi}{3}$
$2)f(x)=tg(2x- \frac{ \pi }{3}), x_0= \frac{ \pi }{6}$

Answer 1

1.f'(x)=5sin^4(x)*cos(x)
f'(pi/3)=5sin^4(pi/3)*cos(pi/3)=5*(sqrt(3)/2)^4*1/2==5*9/16*1/2=45/32
2.f'(x)=2/(cos^2(2x-pi/3))
f'(pi/6)=2/(cos^2(0))=2