Question

x=ln^3t
y=sin(t^2+1)
Найти производную функции

Answer 1

Ответ:

$x = { ln }^{3} (3t) \\ y = \sin( {t}^{2} + 1 )$

$y'x = \frac{y't}{x't}$

$y't = \cos( {t}^{2} + 1 ) \times 2t$

$x't = 3 {ln}^{2} (3t) \times \frac{1}{3t} \times 3 = \\ = \frac{3}{t} {ln}^{2}(3 t)$

$y'x = \frac{2t \cos( {t}^{2} + 1) \times t}{3 {ln}^{2}(3t) } = \\ = \frac{2 {t}^{2} \cos( {t}^{2} + 1) }{3 {ln}^{2}(3 t)}$