## Task Condition

Find the derivative.

$$
y=\frac{5^{x}(\sin 3 x \cdot \ln 5-3 \cos 3 x)}{9+\ln ^{2} 5}
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{5^{x}(\sin 3 x \cdot \ln 5-3 \cos 3 x)}{9+\ln ^{2} 5}\right)^{\prime}= \\
& =\frac{1}{9+\ln ^{2} 5} \cdot\left(5^{x}(\sin 3 x \cdot \ln 5-3 \cos 3 x)\right)^{\prime}= \\
& =\frac{1}{9+\ln ^{2} 5} \cdot\left(5^{x} \cdot \ln 5 \cdot(\sin 3 x \cdot \ln 5-3 \cos 3 x)+5^{x}(\cos 3 x \cdot 3 \cdot \ln 5-3(-\sin 3 x) \cdot 3)\right)= \\
& =\frac{5^{x}}{9+\ln ^{2} 5} \cdot\left(\ln ^{2} 5 \cdot \sin 3 x-3 \ln 5 \cdot \cos 3 x+3 \ln 5 \cdot \cos 3 x+9 \sin 3 x\right)= \\
& =\frac{5^{x}}{9+\ln ^{2} 5} \cdot\left(\ln ^{2} 5 \cdot \sin 3 x+9 \sin 3 x\right)=\frac{5^{x} \cdot \sin 3 x}{9+\ln ^{2} 5} \cdot\left(\ln ^{2} 5+9\right)=5^{x} \cdot \sin 3 x
\end{aligned}
$$

## Problem Kuznetsov Differentiation 15-23