Question

Найдите угол наклона кривой в заданной точке.

Answer 1

tga=f`(x0)
1
f`(x)=(2x³+4x-2x³+4x)/(x²+2)²=8x/9x²+2)²
f`(2)=16/36=4/9
tga=4/9
a=arctg4/9
2
y=1/2*sin8x
f`(x)=1/2*cos8x*8=4cos8x
f`(π/3)=4cos(8π/3)=4cos(2π+2π/3)=4cos(π-π/3)=-4cosπ/3=-4*1/2=-2
tga=-2
a=π-arctg2

Answer 2

tgx=y'

y'=(x²-2)'(x²+2)-(x²-2)(x²+2)' / (x²+2)² =  2x(x²+2)-2x(x²-2)/(x²+2)²=

=2x³+4x-2x³+4x/(x²+2)² = 8x/(x²+2)²  = 8*2/(2²+2)²=16/36 =4/9

tgx=4/9
x=arctg 4/9

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y=sin4xcos4x=1/2 sin8x

y'=1/2 *cos8x * 8=4cos8x=4cos 8π/3 = 4cos (2π+2π/3)=  - 2 четверть ,cos отриц.

=-4cos π/3 = -4*1/2=-2

tgx=-2
x=-arctg2+πn , n∈Z