Найти производную функции при данном значении аргумента 1 f(x)=sin²x, x= п/4 2 F(x)=ln cosx, x= -п/3 3 f(t)=sint-cos²t t=0 4 f(z)=ln tg z, z=п/4 sin x 5 f(x)=l

[latex]1)y=sin^2x[/latex]  [latex]y'=(sin^2x)'=2sinx(sinx)'=2sinx*cosx=sin2x[/latex]  [latex]x=\frac{\pi}4=> y'(\frac{\pi}4)=sin\pi=0[/latex]  [latex]2)y=lncosx[/latex]  [latex]y'=(lncosx)'=\frac{1}{cosx}*(cosx)'=-\frac{sinx}{cosx}=-tgx[/latex]  [latex]x=-\frac{\pi}3=>y'(-\frac{\pi}3)=-tg(-\frac{\pi}3)=tg(\frac{\pi}3)=\sqrt{3}[/latex]  [latex]3) y=sint-cos^2t[/latex]  [latex]y'=(sint-cos^2t)'=(sint)'-(cos^2t)'=cost-2cost(cost)=[/latex]  [latex]=cost+2cost*sint=cost+sin2t[/latex]  [latex]t=0=>y'(0)=cos0-sin0=1-0=1[/latex]  [latex]4) y=lntgz[/latex]  [latex]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}[/latex]  [latex]z=\frac{\pi}4=>y'(\frac{\pi}4)=\frac{1}{sin^2(\frac{\pi}4)}=\frac{1}{\frac{2}{4}}}=\frac{4}2\2[/latex]