Нужно найти производную функции, заданную параметрически.

10-11 класс

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Lyoha123951 19 нояб. 2014 г., 1:11:54 (9 лет назад)

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ZORKHD

19 нояб. 2014 г., 3:05:32 (9 лет назад)

$y'_x= \frac{y'_t}{x'_t}$
$x'_t= \frac{-3cos^2(t)*sin(t) \sqrt{cos(2t)}+ \frac{1}{ \sqrt{cos(2t)} }*sin(2t)*cos^3(t)}{cos(2t)}=\\= \frac{-3cos^2(t)*sin(t)*cos(2t)+2sin(t)*cos^4(t)}{cos(2t) \sqrt{cos(2t)}}=\\= \frac{sin(t)cos^2(t)(-3cos(2t)+2cos^2(t))}{cos(2t) \sqrt{cos(2t)} }= \frac{sin(t)cos^2(t)(3-6cos^2(t)+2cos^2(t))}{cos(2t) \sqrt{cos(2t)} }=\\= \frac{sin(t)cos^2(t)(3-4cos^2(t))}{cos(2t) \sqrt{cos(2t)} }$
$y'_t= \frac{3sin^2(t)*cos(t) \sqrt{cos(2t)}+ \frac{1}{ \sqrt{cos(2t)} }*sin(2t)*sin^3(t)}{cos(2t)} =\\= \frac{3sin^2(t)*cos(t) cos(2t)+2cos(t)*sin^4(t)}{cos(2t) \sqrt{cos(2t)} }= \frac{sin^2(t)*cos(t)*(3cos(2t)+2sin^2(t))}{cos(2t)\sqrt{cos(2t)}}=\\= \frac{sin^2(t)*cos(t)*(3-6sin^2(t)+2sin^2(t))}{cos(2t) \sqrt{cos(2t)} }=\frac{sin^2(t)*cos(t)*(3-4sin^2(t))}{cos(2t) \sqrt{cos(2t)} }$
$y'_x=\frac{sin^2(t)*cos(t)*(3-4sin^2(t))}{sin(t)*cos^2(t)*(3-4cos^2(t))}=\frac{sin(t)*(3-4sin^2(t))}{cos(t)*(3-4cos^2(t))}= \frac{sin(3t)}{-cos(3t)} =-tg(3t)$