Question

Найдите значение производной функции F(x) в точке x0= 0
1) f(x)= sin(x^3+x-pi(пи)/4)
2) f(x)= tg(x^2+pi/6)

Answer 1

1) f(x)= sin(x^3+x-pi/4)
f`(x)=(3x²+1)*cos(x³+x-π/4)
f`(0)=1*cosπ/4=√2/2

2) f(x)= tg(x^2+pi/6)
f`(x)=2x/cos²(x²+π/6)
f`(0)=0/(√3/2)=0

Answer 2

1)f'(x)= cos(x^3+x-pi(пи)/4) * (3x²+1)

f'(0)= 1* cos(0+0-pi(пи)/4 ) =cospi(пи)/4 = √2/2

2) f'(x)= 1/ cos²(x^2+pi/6)  * 2x
f'(0)= 2*0 / cos²(0+pi/6) = 0