1
1)f(x)=∛(x+1)
f(0)=∛1=1
f`(x)=1/(3∛(x+1)²)
f`(0)=1/3
Y=1+1/3*(x-0)=1/3*x+1
2)f(x)=x+1/(x+1)
f(0)=0+1=1
f`(x)=1-1/(x+1)²
f`(0)=1-1=0
Y=1+0(x-0)=1
3)f(x)=sin2x-ln(x+1)
f(0)=sin0-ln1=0-0=0
f`(x)=2cos2x-1/(x+1)
f`(0)=2cos0-1=2*1-1=1
Y=0+1(x-0)=x
2
(1/√(1-4x))`=-1/2√(1-4x)³ *(-4)=2/√(1-4x)³