## Task Condition

Find the derivative.

$$
y=\operatorname{arctg} \frac{2 \sin x}{\sqrt{9 \cos ^{2} x-4}}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\operatorname{arctg} \frac{2 \sin x}{\sqrt{9 \cos ^{2} x-4}}\right)^{\prime}= \\
& =\frac{1}{1+\left(\frac{2 \sin x}{\sqrt{9 \cos ^{2} x-4}}\right)^{2}} \cdot 2 \cdot \frac{\cos x \cdot \sqrt{9 \cos ^{2} x-4}-\sin x \cdot \frac{1}{2 \sqrt{9 \cos ^{2} x-4}} \cdot 18 \cos x \cdot(-\sin x)}{9 \cos ^{2} x-4}= \\
& =\frac{9 \cos ^{2} x-4}{9 \cos ^{2} x-4+4 \sin ^{2} x} \cdot 2 \cdot \frac{\cos x \cdot\left(9 \cos ^{2} x-4\right)+\sin x \cdot 9 \cos x \cdot \sin x}{\left(9 \cos ^{2} x-4\right) \sqrt{9 \cos ^{2} x-4}}= \\
& =\frac{1}{5 \cos ^{2} x} \cdot 2 \cdot \frac{\cos x \cdot\left(9 \cos ^{2} x-4+9 \sin x^{2}\right)}{\sqrt{9 \cos ^{2} x-4}}=\frac{1}{5 \cos x} \cdot 2 \cdot \frac{9-4}{\sqrt{9 \cos ^{2} x-4}}= \\
& =\frac{1}{5 \cos x} \cdot 2 \cdot \frac{5}{\sqrt{9 \cos ^{2} x-4}}=\frac{2}{\cos x \sqrt{9 \cos ^{2} x-4}}
\end{aligned}
$$

Problem Kuznetsov Differentiation 15-17