Răspuns:
a)f(x)=cos 3x
f `(x)= -3sin3x
f `(π/3)=-3sin 3π/3=
-sinπ=0
b)f(x)=ln(2x-3) xo=2
f `(x)=(2x-3)`/(2x-3)=2/(2x-3)
f `(2)=2/(2*2-3)=2/(4-3)=2
c)f(x)=∛x-sin 2x
(∛x) =x^1/3
(x^1/3) `=1/3(x^1/3-1)=1/3x^(-2/3)=
1/3*∛x²
sin2x )`=2cos 2x
f `(x)=1/3*∛x²-2cos2x
f `(1)=1/3*∛1²-2cos 2*1=
1/3*1-2 cos2=
1/3-2cos 2
Explicație pas cu pas:
