## Task Condition

Find the derivative.

$$
y=\ln \frac{\sin x}{\cos x+\sqrt{\cos 2 x}}
$$
## Solution

$y^{\prime}=\left(\ln \frac{\sin x}{\cos x+\sqrt{\cos 2 x}}\right)^{\prime}=\frac{\cos x+\sqrt{\cos 2 x}}{\sin x} \cdot\left(\frac{\sin x}{\cos x+\sqrt{\cos 2 x}}\right)^{\prime}=$

$=\frac{\cos x+\sqrt{\cos 2 x}}{\sin x} \cdot \frac{\cos x \cdot(\cos x+\sqrt{\cos 2 x})-\sin x \cdot\left(-\sin x+\frac{1}{2 \sqrt{\cos 2 x}} \cdot(-2 \sin 2 x)\right)}{(\cos x+\sqrt{\cos 2 x})^{2}}=$

$=\frac{\cos ^{2} x \cdot \sqrt{\cos 2 x}+\cos x \cdot \cos 2 x+\sin ^{2} x \cdot \sqrt{\cos 2 x}+\sin 2 x \cdot \sin x}{\sin x \cdot(\cos x+\sqrt{\cos 2 x}) \cdot \sqrt{\cos 2 x}}=$

$=\frac{\sqrt{\cos 2 x} \cdot\left(\cos ^{2} x+\sin ^{2} x\right)+\cos x \cdot(1-2 \sin 2 x)+2 \sin ^{2} x \cdot \cos x}{\sin x \cdot(\cos x+\sqrt{\cos 2 x}) \cdot \sqrt{\cos 2 x}}=$

$=\frac{\sqrt{\cos 2 x}+\cos x \cdot\left(1-2 \sin ^{2} x+2 \sin ^{2} x\right)}{\sin x \cdot(\cos x+\sqrt{\cos 2 x}) \cdot \sqrt{\cos 2 x}}=\frac{\sqrt{\cos 2 x}+\cos x}{\sin x \cdot(\cos x+\sqrt{\cos 2 x}) \cdot \sqrt{\cos 2 x}}=$

$=\frac{1}{\sin x \cdot \sqrt{\cos 2 x}}$

Problem Kuznetsov Differentiation 15-8