Časť:
Project ID:
9000063104
Accepted:
1
Derivácia funkcie \(f\colon y = \frac{\sin x}
{\sin x-\cos x}\)
je rovná:
\(f'(x) = \frac{-1}
{(\sin x-\cos x)^{2}} ,\ x\neq \frac{\pi }{4} + k\pi ;k\in \mathbb{Z}\)
\(f'(x) = \frac{\sin ^{2}x-\cos ^{2}x}
{(\sin x-\cos x)^{2}} ,\ x\neq \frac{\pi }{4} + k\pi ;k\in \mathbb{Z}\)
\(f'(x) = \frac{\sin x(\cos x+1)}
{(\sin x-\cos x)^{2}} ,\ x\neq \frac{\pi }{4} + k\pi ;k\in \mathbb{Z}\)
\(f'(x) = \frac{\cos ^{2}x-\sin ^{2}x}
{(\sin x-\cos x)^{2}} ,\ x\neq \frac{\pi }{4} + k\pi ;k\in \mathbb{Z}\)