## Task Condition

Find the derivative.

$y=\frac{7^{x}(3 \sin 3 x+\cos 3 x \cdot \ln 7)}{9+\ln ^{2} 7}$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\frac{7^{x}(3 \sin 3 x+\cos 3 x \cdot \ln 7)}{9+\ln ^{2} 7}\right)^{\prime}= \\
& =\frac{1}{9+\ln ^{2} 7} \cdot\left(7^{x}(3 \sin 3 x+\cos 3 x \cdot \ln 7)\right)^{\prime}= \\
& =\frac{1}{9+\ln ^{2} 7} \cdot\left(7^{x} \cdot \ln 7 \cdot(3 \sin 3 x+\cos 3 x \cdot \ln 7)+7^{x}(3 \cos 3 x \cdot 3-\sin 3 x \cdot 3 \cdot \ln 7)\right)= \\
& =\frac{7^{x}}{9+\ln ^{2} 7} \cdot\left(3 \ln 7 \cdot \sin 3 x+\cos 3 x \cdot \ln ^{2} 7+9 \cos 3 x-3 \ln 7 \cdot \sin 3 x\right)= \\
& =\frac{7^{x} \cdot \cos 3 x}{9+\ln ^{2} 7} \cdot\left(\ln ^{2} 7+9\right)=7^{x} \cdot \cos 3 x
\end{aligned}
$$

## Problem Kuznetsov Differentiation 15-7