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 $$(3s - 15)/(2s^2 -4s + 10)$$
For c),
$$
\frac{{3s - 15}}{{2s^2  - 4s + 10}} = \frac{3}{2}\frac{{s - 5}}{{s^2  - 2s + 5}} = \frac{3}{2}\frac{{s - 1 - 4}}{{(s - 1)^2  + 4}} = \frac{3}{2}\frac{{s - 1}}{{(s - 1)^2  + 2^2}} - 3\frac{2}{{(s - 1)^2  + 2^2 }}.
$$
Now let $\alpha = -1$ and $\omega = 2$, and consider 8-9 of this table.

EDIT: By 9 and 8 of that table, the inverse transforms of $\frac{{s + \alpha }}{{(s + \alpha )^2  + \omega ^2 }}$ and  $\frac{\omega}{{(s + \alpha)^2  + \omega^2 }}$ are $e^{-\alpha t} \cos(\omega t)$ and $e^{-\alpha t} \sin (\omega t)$, respectively. Hence the inverse transform of $\frac{{3s - 15}}{{2s^2  - 4s + 10}}$ is 
$$
\frac{3}{2}e^t \cos (2t) - 3e^t \sin (2t).
$$
I'm going to GUESS that in (c), you really meant $(3s-15)/(2s^2-4s-10)$, despite the fact that you wrote $3s-\Big(15/(2s^2-4s-10)\Big)$.

The quadratic polynomial $2s^2-4s-10$ is $0$ when $s=1\pm\sqrt{6}$, so it factors as $2(s-1-\sqrt{6})(s-1+\sqrt{6})$.  Then you can do a partial-fraction decomposition as
$$
\frac{3s-15}{2s^2-4s-10} = \frac{A}{s-1-\sqrt{6}} + \frac{B}{s-1+\sqrt{6}}.
$$
You need to figure out numbers what $A$ and $B$ are.

Then find the inverse Laplace transform of each of the two terms separately.

Later edit:  Oh.  I see you had $+10$, not $-10$.  So the quadratic is irreducible.  But you should still find it useful to bear this kind of thing in mind.  I see that two others went through partial fractions without mentioning the term "partial fractions".  And they completed the square without mentioning the term "completing the square".