assuming you mean
[tex]f(x)=(x+1)^5-5x-2[/tex]
first use chain rule on first part
derivitive of (x+1)⁵=5(x+1)⁴ times the derivitive of (x+1) which is 1
the first part is 5(x+1)⁴
derivitive of -5x is -5
derivitive of -2 is 0
we gots
f'(x)=5(x+1)⁴-5
find zeroes
0=5(x+1)⁴-5
5=5(x+1)⁴
1=(x+1)⁴
+/-1=x+1
-1+/-1=x
-2=x or 0=x
f'(x)=5(x+1)⁴-5
zeroes at x=-2 and x=0
