Mathematic Problems

257. példa

y = \ln \frac{\sqrt{2 x + 1}}{\sin x}

y = \ln \frac{\sqrt{2 x + 1}}{\sin x}

y = \ln \frac{{(2 x + 1)}^{\frac{1}{2}}}{\sin x}

y' = \frac{\sin x}{{(2 x + 1)}^{\frac{1}{2}}} \cdot \frac{\frac{1}{2} {(2 x + 1)}^{- \frac{1}{2}} \cdot 2 \sin x - {(2 x + 1)}^{\frac{1}{2}} \cdot \cos x}{\sin^{2} x}

= {(2 x + 1)}^{\frac{1}{2}}

= \frac{\frac{\sin x}{\sqrt{2 x + 1}} - \sqrt{2 x + 1} \cos x}{\sqrt{2 x + 1} \sin x}

= \frac{\sin x - (2 x + 1) \cos x}{(2 x + 1) \sin x}

= \frac{1}{2 x + 1} - c t g x

y' = \frac{1}{2 x + 1} - c t g x

Matematika - Képlettár