Найти производную функции при данном значении аргумента 1 f(x)=sin²x, при x= п/4. 2...

$1)y=sin^2x$

$y'=(sin^2x)'=2sinx(sinx)'=2sinx*cosx=sin2x$

y'(\frac{\pi}4)=sin\pi=0" alt="x=\frac{\pi}4=> y'(\frac{\pi}4)=sin\pi=0" align="absmiddle" class="latex-formula">

$2)y=lncosx$

$y'=(lncosx)'=\frac{1}{cosx}*(cosx)'=-\frac{sinx}{cosx}=-tgx$

y'(-\frac{\pi}3)=-tg(-\frac{\pi}3)=tg(\frac{\pi}3)=\sqrt{3}" alt="x=-\frac{\pi}3=>y'(-\frac{\pi}3)=-tg(-\frac{\pi}3)=tg(\frac{\pi}3)=\sqrt{3}" align="absmiddle" class="latex-formula">

$3) y=sint-cos^2t$

$y'=(sint-cos^2t)'=(sint)'-(cos^2t)'=cost-2cost(cost)=$

$=cost+2cost*sint=cost+sin2t$

y'(0)=cos0-sin0=1-0=1" alt="t=0=>y'(0)=cos0-sin0=1-0=1" align="absmiddle" class="latex-formula">

$4) y=lntgz$

$y'=(lntgz)'=\frac{1}{tgz}(tgz)'=\frac{\frac{1}{cos^2z}}{tgz}=\frac{\frac{1}{cos^2z}}{\frac{sin^2x}{cos^2x}}=\frac{cos^2x}{sin^2xcos^2x}=\frac{1}{sin^2x}$

y'(\frac{\pi}4)=\frac{1}{sin^2(\frac{\pi}4)}=\frac{1}{\frac{2}{4}}}=\frac{4}2\2" alt="z=\frac{\pi}4=>y'(\frac{\pi}4)=\frac{1}{sin^2(\frac{\pi}4)}=\frac{1}{\frac{2}{4}}}=\frac{4}2\2" align="absmiddle" class="latex-formula">