Časť:
Project ID:
9000063303
Accepted:
1
Derivácia funkcie \(f\colon y = \sqrt{\sin x}\)
je rovná:
\(f'(x) = \frac{\cos x}
{2\sqrt{\sin x}},\ x\in \mathop{\mathop{\bigcup
}}\nolimits _{k\in \mathbb{Z}}\left (2k\pi ;\pi + 2k\pi \right )\)
\(f'(x) = \frac{\sin x}
{2\sqrt{\cos x}},\ x\in \mathop{\mathop{\bigcup
}}\nolimits _{k\in \mathbb{Z}}\left (2k\pi ; \frac{\pi } {2} + 2k\pi \right )\)
\(f'(x) = \frac{1}
{2\sqrt{\sin x}},\ x\in \mathop{\mathop{\bigcup
}}\nolimits _{k\in \mathbb{Z}}\left (2k\pi ;\pi + 2k\pi \right )\)
\(f'(x) = \frac{\cos x}
{2\sqrt{\sin x}},\ x\in \mathop{\mathop{\bigcup
}}\nolimits _{k\in \mathbb{Z}}\left \langle 2k\pi ; \frac{\pi } {2} + 2k\pi \right \rangle \)