Calculus

f(s) = √(s^2 + 1) / (s^2 + 4)
f'(s) = [(s^2 + 4)(d/dx)√(s^2 + 1) - √(s^2 + 1)(d/dx)(s^2 + 4)] / (s^2 + 4)^2
= {[(s^2 + 4)(2s)/[2√(s^2 + 1)] - √(s^2 + 1)(2s)]} / (s^2 + 4)^2
= [(s^2 + 4)(s)/√(s^2 + 1) - 2s√(s^2 + 1)] / (s^2 + 4)^2
= [(s^2 + 4)(s) - 2s(s^2 + 1)] / [(s^2 + 4)^2 √(s^2 + 1)]
= (2s - s^3) / [(s^2 + 4)^2 √(s^2 + 1)]

y = 2^(3x)^(2)
√y = 2^(3x)
ln √y = ln 2^(3x)
1/2 ln y = 3x ln 2
(1/2)(1/y)(dy/dx) = 3ln 2
dy/dx = 6yln 2 = (6ln 2)[2^(3x)^(2)]

For qusetion 2 √(2^(3x)^(2)) can be either + 2^(3x) or- 2^(3x)
Therefore y = 2^(3x)^(2)
y=2^(6x)
ln y = 6x ln 2
1/y dy/dx = 6ln2
dy/dx = 6yln2 = (6ln2)(2^(3x)^(2))

Hi, this is Question 1
http://postimg.org/image/ig9xzec9z/
Sorry for the confusion

q1
係
[ √ ( s^2+1 ) ] / (s^2 +4 )
or
√ [ (s^2+1) / (s^2 +4 ) ]
or
√ { s^2+ [ 1 / (s^2 +4 ) ] } ?