Question

Ajutor urgent!!!
Sa se scrie ecuatia tangentei la graficul functie $f:R-R$ , $f(x)= \frac{1}{2} sin^{2} (4x- \frac{ \pi }{3} )$ in punctul cu abcisa $x_{0}= \frac{ \pi }{6}$ 
ecuatia tangentei este: $y= f ^{'}( x_{0})(x- x_{0}) + f( x_{0} )$  si raspunsul trebuie sa fie $y= \sqrt{3} x - \frac{ \pi \sqrt{3} }{6} + \frac{3}{8}$

Answer 1

Ecuatia tangentei in punctul $M(x_{0};y_{0})$ este:
[tex]y-y_{0}=f'(x_0{})(x-x_{0})\\ y_{0}=f(\frac{\pi}{6})=\frac{1}{2}sin^2(\frac{\pi}{3})=\frac{1}{2}\cdot \frac{3}{4}=\frac{3}{8}\\ f'(x)=\frac{1}{2}\cdot 2sin(4x-\frac{\pi}{3})\cdot cos(4x-\frac{\pi}{3}) \cdot 4\\ f'(\frac{\pi}{3})=4\cdot sin\frac{\pi}{3}\cdot cos \frac{\pi}{3}=\sqrt{3}\\ y-\frac{3}{8}=\sqrt{3}(x-\frac{\pi}{6})\\ y=\sqrt{3}x-\frac{\pi\sqrt{3}}{6}+\frac{3}{8}[/tex]