Вычислите производную:
1) f (t)=2lg (t+1) , t=1
2) f(x)= -ln cosx , x=pi/4

1)  f ^{'}(t) = \frac{2}{(t+1) ln10} f ^{'}(1) = \frac{2}{(1+1) ln10} = \frac{1}{ln10}2)  f^{'}(x) = - \frac{1}{cos x} * (-sin x) = tg x f^{'}( \frac{ \pi }{4} ) = tg \frac{ \pi }{4} = 1