Could someone please help me find the derivatives to these 2 problems?

Question

1. y= 2sin x-tan x

2. y= x sec x

Answer 1

1. y' = 2cos x - sec^2 x
2. y' = x*[tan x]*[sec x] + sec x

Answer 2

1. y'=2cosx-sec^(2)x
2. I would have to look at my notes... sec's derivative is a fraction I think

Answer 3

y = 2sin(x) - tan(x)

d/dx [ f(x) - g(x) ] = d/dx [f(x)] - d/dx [g(x)]

y' = 2 d/dx sin(x) - d/dx tan(x)

derivative of sin(x) is cos(x)
derivative of tan(x) is sec^2 (x)

y' = 2cos(x) - sec^2 (x)


2) product rule
d/dx (uv) = u'v + v'u

y' = d/dx(x) sec(x) + d/dx sec(x) * x

derivative of sec(x) is sec(x) tan(x)

y' = sec(x) + x sec(x) tan(x)

OR

y' = sec(x) (1 + x tan(x) )