Question

tolong bro jawab pertanyaannya susah bet anjir

Answer 1

Jawaban:

Note: Koreksi jika salah atau jawaban kurang memuaskan

Answer 2

Jawab:

A

Penjelasan dengan langkah-langkah:

Turunan fungsi perkalian f(x) = uv → f '(x) = u'v + uv'

[tex]\begin{matrix}\begin{aligned}u&=3x\\u'&=3\end{aligned} & \begin{aligned}v&=\cot x\\v'&=-\csc^2 x\end{aligned}\end{matrix}\\\begin{aligned}f(x)&=3x\cot x\\f'(x)&=3\cot x+3x(-\csc^2 x)\\f'(x)&=3(\cot x-x\csc^2 x)\\f'\left ( \frac{\pi}{3} \right )&=3\left ( \cot\frac{\pi}{3}-\frac{\pi}{3}\csc^2\frac{\pi}{3} \right )\\m&=3\left [ \frac{\sqrt{3}}{3}-\frac{\pi}{3}\left ( \frac{2\sqrt{3}}{3} \right )^2 \right ]\\&=\sqrt{3}-\frac{4\pi}{3}\end{aligned}[/tex]

Cari ordinat nya

[tex]\begin{aligned}y&=3x\cot x\\&=3\cdot\frac{\pi}{3}\cot\frac{\pi}{3}\\&=\frac{\sqrt{3}}{3}\pi\end{aligned}[/tex]

Persamaan garis singgung nya

[tex]\begin{aligned}y-y_1&=m(x-x_1)\\y-\frac{\pi\sqrt{3}}{3}&=\left ( \sqrt{3}-\frac{4\pi}{3} \right )\left ( x-\frac{\pi}{3} \right )\\y&=x\sqrt{3}-\frac{\pi\sqrt{3}}{3}-\frac{4\pi}{3}x+\frac{4\pi^2}{9}+\frac{\pi\sqrt{3}}{3}\\y&=\left ( \sqrt{3}-\frac{4\pi}{3} \right )x+\frac{4\pi^2}{9}\end{aligned}[/tex]