## Task Condition

Find the derivative.

$$
y=\frac{3^{x}(4 \sin 4 x+\ln 3 \cdot \cos 4 x)}{16+\ln ^{2} 3}
$$
## Solution

$y^{\prime}=\left(\frac{3^{x}(4 \sin 4 x+\ln 3 \cdot \cos 4 x)}{16+\ln ^{2} 3}\right)^{\prime}=$

$=\frac{1}{16+\ln ^{2} 3} \cdot\left(3^{x} \cdot \ln 3 \cdot(4 \sin 4 x+\ln 3 \cdot \cos 4 x)+3^{x}(16 \cos 4 x-4 \ln 3 \cdot \sin 4 x)\right)=$ $=\frac{3^{x}}{16+\ln ^{2} 3} \cdot\left(4 \ln 3 \cdot \sin 4 x+\ln ^{2} 3 \cdot \cos 4 x+16 \cos 4 x-4 \ln 3 \cdot \sin 4 x\right)=$

$=\frac{3^{x}}{16+\ln ^{2} 3} \cdot\left(\ln ^{2} 3 \cdot \cos 4 x+16 \cos 4 x\right)=\frac{3^{x} \cdot \cos 4 x}{16+\ln ^{2} 3} \cdot\left(\ln ^{2} 3+16\right)=3^{x} \cdot \cos 4 x$

## Problem Kuznetsov Differentiation 15-20