9000063307
Parte:
C
Deriva la siguiente función.
\[
f(x) =\ln\left(\cos 2x\right)
\]
\(f'(x) = -2\mathop{\mathrm{tg}}\nolimits 2x,\ x\in \mathop{\mathop{\bigcup
}}\nolimits _{k\in \mathbb{Z}}\left (-\frac{\pi }{4} + k\pi ; \frac{\pi } {4} + k\pi \right )\)
\(f'(x) = 2\mathop{\mathrm{tg}}\nolimits 2x,\ x\in \mathop{\mathop{\bigcup
}}\nolimits _{k\in \mathbb{Z}}\left (-\frac{\pi }{4} + k\pi ; \frac{\pi } {4} + k\pi \right )\)
\(f'(x) = -2,\ x\in \mathop{\mathop{\bigcup
}}\nolimits _{k\in \mathbb{Z}}\left (-\frac{\pi }{4} + k\pi ; \frac{\pi } {4} + k\pi \right )\)
\(f'(x) = 1 -\ln\left(\sin 2x\right),\ x\in \mathop{\mathop{\bigcup
}}\nolimits _{k\in \mathbb{Z}}\left (k\pi ; \frac{\pi } {2} + k\pi \right )\)