Решение
<span>Найти производную
1) у=sin3х -cos3х и вычислить ее значение, если х=3П/4;
y` = 3cos3x + 3sin3x
</span><span>если х=3П/4
</span>y`(3π/4) = 3cos(3*3π/4) + 3sin(3*3π/4) = 3cos(2π + π/4) + 3sin(2π + π/4) =
= 3cosπ/4 + 3sinπ/4 = 3√2/2 + 3√2/2 = 6√2/2 = 3<span>
2) у=</span>√(<span>25-9х), если х=1
y` = - 9 / [2</span>√(25 - 9x)
y`(1) = - 9 / [2<span>√(25 - 9*1) = - 9 / (2</span>√16) = - 9 / 8 = - 1(1/8)<span>
</span>
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3x²-10x+3=3(x-1/3)(x-3)=(3x-1)(x-3)
D=100-36=64
x1=(10-8)/6=1/3
x2=(10+8)/6=3
(3x²-10x+3)/(x²-3x)=(3x-1)(x-3)/[x(x-3)]=(3x-1)/x
(sina-sin3a-sin5a-sin7a)/(cosa-cos3a+cos5a-cos7a)
5a-(7-2*(3-a)-3)=5a-(7-6+2a-3)=
=5a-(2a-2)=5a-2a+2=3a+2
Ответ : 3a+2