## Task Condition

Find the derivative.

$$
y=\frac{2^{x}(\sin x+\cos x \cdot \ln 2)}{1+\ln ^{2} 2}
$$
## Solution

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

## Problem Kuznetsov Differentiation 15-25