Question

f'(x)=2^x/sinx помогите найти производную

Answer 1

$F'(x)= \displaystyle \bigg( \frac{2^x}{sinx} \bigg)' = \displaystyle \frac{(2^x)'*sinx-2^x*(sinx)'}{sin^2x} = \\ \\ \\ = \displaystyle \frac{2^x*ln2*sinx-2^x*cosx}{sin^2x}$