Question

Знайти похідну функції:
y=x³tgx

Answer 1

$y = {x}^{3} \tg(x) \\ y' =( {x}^{3} )' \tg(x) + {x}^{3} ( \tg(x) )' = \\ = 3 {x}^{3 - 1} \tg(x) + {x}^{3} \times \frac{1}{ \cos {}^{2} (x) } = \\ = 3 {x}^{2} \tg(x) + \frac{ {x}^{3} }{ \cos {}^{2} (x) } = \frac{3 {x}^{2} \tg(x) \cos {}^{2} (x) + {x}^{3} }{ \cos {}^{2} (x) } = \\ = \frac{3 {x}^{2} \frac{ \sin(x) \cos {}^{2} (x) }{ \cos (x) } + {x}^{3} }{ \cos {}^{2} (x) } = \frac{3 {x}^{2} \sin(x) \cos(x) + {x}^{3} }{ \cos {}^{2} (x) } = \\ = \frac{2 \times 3 {x}^{2} \sin(x) \cos(x) + 2{x}^{3} }{2 \cos {}^{2} (x) } = \frac{3 {x}^{2} \sin(2x) + 2 {x}^{3} }{ \cos {}^{2} (x) } = \\ = \frac{ {x}^{2}(3 \sin(2x) + 2x) }{ \cos {}^{2} (x) }$