## Task Condition

Find the derivative.

$$
y=\frac{5^{x}(2 \sin 2 x+\cos 2 x \cdot \ln 5)}{4+\ln ^{2} 5}
$$
## Solution

$y^{\prime}=\left(\frac{5^{x}(2 \sin 2 x+\cos 2 x \cdot \ln 5)}{4+\ln ^{2} 5}\right)^{\prime}=$

$$
\begin{aligned}
& =\frac{1}{4+\ln ^{2} 5} \cdot\left(5^{x} \cdot \ln 5 \cdot(2 \sin 2 x+\cos 2 x \cdot \ln 5)+5^{x} \cdot(4 \cos 2 x-2 \sin 2 x \cdot \ln 5)\right)= \\
& =\frac{5^{x}}{4+\ln ^{2} 5} \cdot\left(2 \ln 5 \cdot \sin 2 x+\ln ^{2} 5 \cdot \cos 2 x+4 \cos 2 x-2 \ln 5 \sin 2 x\right)= \\
& =\frac{5^{x}}{4+\ln ^{2} 5} \cdot\left(\ln ^{2} 5 \cdot \cos 2 x+4 \cos 2 x\right)=\frac{5^{x} \cdot \cos 2 x}{4+\ln ^{2} 5} \cdot\left(\ln ^{2} 5+4\right)=5^{x} \cdot \cos 2 x
\end{aligned}
$$

## Problem Kuznetsov Differentiation 15-18