## Task Condition

Find the derivative.

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

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

$=\frac{1}{16+\ln ^{2} 4} \cdot\left(4^{x} \cdot \ln 4 \cdot(\ln 4 \cdot \sin 4 x-4 \cos 4 x)+4^{x}(4 \ln 4 \cdot \cos 4 x+16 \sin 4 x)\right)=$

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

## Problem Kuznetsov Differentiation 15-21