## Task Condition

Find the derivative.

$$
y=\frac{6^{x}(\sin 4 x \cdot \ln 6-4 \cos 4 x)}{16+\ln ^{2} 6}
$$
## Solution

$y^{\prime}=\left(\frac{6^{x}(\sin 4 x \cdot \ln 6-4 \cos 4 x)}{16+\ln ^{2} 6}\right)^{\prime}=\left(\frac{6^{x} \sin 4 x \cdot \ln 6-4 \cdot 6^{x} \cos 4 x}{16+\ln ^{2} 6}\right)^{\prime}=$

$$
=\frac{\ln 6\left(6^{x} \sin 4 x\right)^{\prime}-4\left(6^{x} \cos 4 x\right)^{\prime}}{16+\ln ^{2} 6}=
$$

$$
\begin{aligned}
& =\frac{\ln 6\left(6^{x} \ln 6 \cdot \sin 4 x+6^{x} \cdot \cos 4 x \cdot 4\right)-4\left(6^{x} \ln 6 \cdot \cos 4 x-6^{x} \cdot \sin 4 x \cdot 4\right)}{16+\ln ^{2} 6}= \\
& =\frac{6^{x} \ln ^{2} 6 \cdot \sin 4 x+6^{x} \ln 6 \cdot \cos 4 x \cdot 4-6^{x} \ln 6 \cdot \cos 4 x \cdot 4+6^{x} \cdot \sin 4 x \cdot 16}{16+\ln ^{2} 6}= \\
& =\frac{\sin 4 x \cdot 6^{x}\left(16+\ln ^{2} 6\right)}{16+\ln ^{2} 6}=6^{x} \sin 4 x
\end{aligned}
$$