Question

Помогите, пожалуйста(・・?)

Answer 1

$y'=(-3\sin{3x})'=-3\cdot\cos{3x}\cdot (3x)'=-3\cdot 3 \cdot \cos{3x}=-9\cos{3x}$

Уравнение касательной:

$y-f(x_0)=f'(x_0)\cdot (x-x_0)$

$f(x_0)=-3\sin{(\frac{3\pi}{4})}=-3\sin{(\pi-\frac{\pi}{4})}=-3\cdot \frac{\sqrt{2}}{2}=-\frac{3\sqrt{2}}{2} \\ \\ f'(x_0)=-9\cos{(\frac{3\pi}{4})}=-9\cos{(\pi-\frac{\pi}{4})}=-9\cdot (-\frac{\sqrt{2}}{2})=\frac{9\sqrt{2}}{2} \\ \\ y-(-\frac{3\sqrt{2}}{2})=\frac{9\sqrt{2}}{2}\cdot(x-\frac{\pi}{4}); \\ \\ y+\frac{3\sqrt{2}}{2}=\frac{9\sqrt{2}\cdot x }{2}-\frac{9\pi\sqrt{2}}{8}; \\ \\ y=\frac{9\sqrt{2}\cdot x }{2}-\frac{9\pi\sqrt{2}}{8}-\frac{3\sqrt{2}}{2};$

$y=\frac{3\sqrt{2}}{2}\cdot(3x-\frac{3\pi}{4}-1)$