Question

Найти производную сложной функции

y= sin ln^3 (2x+1)

Answer 1

$f'(x)\cos\left(\ln^{3}\left(2\,x+1\right)\right)\cdot \left(\ln^{3}\left(2\,x+1\right)\right)'=$

$=\cos\left(\ln^{3}\left(2\,x+1\right)\right)\cdot 3\cdot \ln^{2}\left(2\,x+1\right)\cdot \left(\ln\left(2\,x+1\right)\right)'=$

$=3\,\ln^{2}\left(2\,x+1\right)\,\cos\left(\ln^{3}\left(2\,x+1\right)\right)\cdot \dfrac{1}{2\,x+1}\cdot \left(2\,x+1\right)$

$=\dfrac{3\,\ln^{2}\left(2\,x+1\right)\,\cos\left(\ln^{3}\left(2\,x+1\right)\right)}{2\,x+1}\cdot \left(2\cdot 1+0\right)=\dfrac{6\,\ln^{2}\left(2\,x+1\right)\,\cos\left(\ln^{3}\left(2\,x+1\right)\right)}{2\,x+1}$