<s>[INST] Laplace transform and Differentiation help I need to know how to differentiate something like these 
a) Differentiate with respect to s  $$ b/(s^2 +b^2)$$
b) Solve the given equation 
                               $$f(t) = s(1/s^2 + 2/s^3)$$
c) Reverse Laplace transform of $$(3s - 15)/(2s^2 -4s + 10)$$
 [/INST] As for (c), first of all, get rid of that $2$ factor in the denominator: $2s^2 -4s +10 = 2(s^2 -2s +5)$ and observe that the polynomial $s^2 -2s +5$ has no real roots. So, we write it as follows:
$$
s^2 -2s + 5 = (s-1)^2 + (5 - 1) = (s-1)^2 + 2^2
$$
Next, we try to decompose our fraction like this:
$$
\frac{3s-15}{(s-1)^2 + 2^2} = A \frac{s-1}{(s-1)^2 + 2^2} + B \frac{2}{(s-1)^2 + 2^2} \ ,
$$
for some constants $A$ and $B$ which you'll have no trouble to find. 
Why do we do this? Well, because we have listen to what our professor said in classroom and we know that each member on the right-hand side has an easy inverse Laplace transform.   :-)
Namely,
$$
{\cal L}^{-1}\left\{\frac{1}{2} \frac{3s-15}{(s-1)^2 + 2^2} \right\} = \frac{A}{2} {\cal L}^{-1} \left\{ \frac{s-1}{(s-1)^2 + 2^2} \right\} + \frac{B}{2} {\cal L}^{-1} \left\{ \frac{2}{(s-1)^2 + 2^2} \right\}
$$
Which means
$$
{\cal L}^{-1}\left\{\frac{1}{2} \frac{3s-15}{(s-1)^2 + 2^2} \right\} = \frac{A}{2} e^t \cos (2t) + \frac{B}{2} e^t \sin (2t)
$$
</s>