Derivatives help in AP calculus

Can anyone help me do these problems?? My teacher sent us the answer key and i did them wrong and i'd like to know what i did wrong.
1. y=8
2. y=x^6
3. y-=1/x^7
4. y=5 sqrt. X
5. f(x)= x+1
6. f(x)=-2x^2+3x-6
7. f(x)=x^2+ 4x^3
8. f(t)=t^3-2t+ 4
9. y=pi/2 sin-cos
10. y=x^2-1/2cosx
11. y=1/x-3 sinx
12. f(x)= x^3 tanx
13. f(x)= (x^2+1)(x^2-2x)
14. f(x)= 5x sec x+ tan x
15. f(x)= (x+1/x+2)(2x-5)

1.) dy/dx = 0 ... Derivative of a constant (with respect to x) is 0.

2.) dy/dx = 6x^5 ... Consult the formula for the derivative of a polynomial.

3.) y = 1/x^7 = x^(-7)

dy/dx = -7x^(-8)

dy/dx = -7/x^8

4.) y = 5√x = 5x^(1/2)

dy/dx = 5(1/2)x^(1/2-1) = (5/2)x^(-1/2)

dy/dx = 5/2√x

5.) f '(x) = 1

6.) f '(x) = -4x + 3

7.) f '(x) = 2x + 12x^2

8.) f '(t) = 3t^2 -2

9.) I have no idea. Poor syntax. Please rewrite.

10.) Confusing syntax. Judging by the caliber of these problems I am assuming you mean:

y = x^2 - (1/2) cos x

dy/dx = 2x - (1/2)(-sin x)

dy/dx = 2x + (1/2) sin x

11.) y = 1/x - 3 sin x

y = x^(-1) - 3 sin x

dy/dx = (-1)x^(-2) - 3 cos x

dy/dx = -1/x^2 - 3 cos x

12.) f(x) = x^3 tan x

f '(x) = 3x^2 tan x + x^3 sec^2 x ... The product rule.

13.) f(x) = (x^2 + 1)(x^2 - 2x)

f(x) = x^4 - 2x^3 + x^2 - 2x

f '(x) = 4x^3 - 6x^2 + 2x -2

14.) f(x) = 5x sec x + tan x

f '(x) = 5 sec x + 5x sec x tan x + sec^2 x

15.) f(x) = [ (x+1)/(x+2) ](2x-5)

f(x) = (2x^2 - 5x + 2x - 5)/(x + 2)

f(x) = (2x^2 - 3x - 5)/(x + 2)

f '(x) = [ (x + 2)(4x - 3) - (2x^2 - 3x - 5)(1) ] / (x + 2)^2 ... The quotient rule.

f '(x) = [ (x + 2)(4x - 3) - (2x^2 - 3x - 5) ] / (x + 2)^2

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