y=xSin( √ 2x-1) + log2 (3x+1)

We need the product and chain rules for the first piece.
(x)′ = 1
( sin (x) )′ = cos (x)
( √[x] )′ = 1 / 2√[x]
( 2x-1 )′ = 2

So, sin( √ 2x-1)′ = cos(√[2x - 1]) * (1 / 2√[2x - 1]) * 2
= cos(√[2x - 1]) / √[2x-1]

Making the derivative of the first chunk = x{ cos(√[2x - 1]) / √[2x-1] } + sin( √ 2x-1)

Now for the second chunk, we just need the chain rule, but first we need to rewrite the logarithm:

log2 (3x+1) = ln(3x+1)/ln(2) = [1/ln(2)] * ln(3x+1)
( ln(3x+1) )′ = 3 / (3x+1)

The whole nasty first derivative is x{ 2cos(√[2x - 1]) / √[2x-1] } + sin( √ 2x-1) + 3 / [ ln(2)(3x+1) ]

The second derivative requires application of the chain, product, and quotient rules and should be quite awful. I'm sorry I don't have time right now to work it out for you.

y={1+cos(1-x)/e^3x} + ln(2x+5)

Same approach as before. First part:
( cos(1-x) )′ = sin(1-x)
(e^3x)′ = 3e^3x

1 + { (e^3x)(sin(1-x)) - (cos(1-x))(3e^3x) } / e^6x
factor out an e^3x:
1 + (e^3x/e^3x) { sin(1-x) - 3cos(1-x) } / e^3x
= 1 + { sin(1-x) - 3cos(1-x) } / e^3x

Second part:
(ln(2x+5))′ = 2 / (2x + 5)

Put them together:
1 + { sin(1-x) - 3cos(1-x) } / e^3x + 2 / (2x + 5)

Again, sorry I don't have time for ƒ″. I hope I helped point you in the right direction.

Don't thank me too soon mike, I just realized I made a mistake!

In the second problem, I factored e^3x out of e^6x and put the answer as e^3x. But it's e^2x!

Sorry! I should have caught that...