Question

надо найти производную функции,заранее спасибо

Answer 1

Производная функции, заданной параметрически -  $\displaystyle\\\frac{y'_t}{x'_t} =y'_x$

$\displaystyle\\x=\frac{a}{3}(2\cos(t)+\cos(2t))\\\\x'_t=\bigg( \frac{a}{3}(2\cos(t)+\cos(2t)) \bigg)'_t=\frac{a}{3}\bigg(2\cos(t)+\cos(2t) \bigg)'_t=\frac{a}{3}*\bigg( 2*(-\sin(t))-\\\\\\-\sin(2t)*2\bigg)=-\frac{2a*\sin(t)+2a*\sin(2t)}{3}\\\\\\y'_t=\bigg(\frac{a}{3}(2\sin(t)-\sin(2t)) \bigg)'_t=\frac{a}{3}\bigg(2\sin(t)-\sin(2t) \bigg)'_t =\frac{a}{3}\bigg( 2\cos(t)-2\cos(2t)\bigg)=\\\\\\=\frac{2a\cos(t)-2a\cos(2t)}{3}$

$\displaystyle\\y'_x=\frac{y'_t}{x'_t}\Rightarrow \frac{\dfrac{2a\cos(t)-2a\cos(2t)}{3} }{-\dfrac{2a\sin(t)+2a\sin(2t)}{3} } =-\frac{2a(\cos(t)-\cos(2t))}{2a(\sin(t)+\sin(2t))}=\\\\\\=\frac{-2\sin\bigg(\dfrac{3t}{2}\bigg)\sin\bigg(-\dfrac{t}{2}\bigg) }{2\sin\bigg(\dfrac{3t}{2}\bigg)\cos\bigg(-\dfrac{t}{2}\bigg) }=-\frac{\sin\bigg(\dfrac{t}{2} \bigg)}{\cos\bigg(\dfrac{t}{2} \bigg)}=-tg\bigg(\frac{t}{2}\bigg)$