A tough differential calculus problem This is a question I've had a lot of trouble with. I HAVE solved it, however, with a lot of trouble and with an extremely ugly calculation. So I want to ask you guys (who are probably more 'mathematically-minded' so to say) how you would solve this. Keep in mind that you shouldn't use too advanced stuff, no differential equations or similair things learned in college:
Given are the functions $f_p(x) = \dfrac{9\sqrt{x^2+p}}{x^2+2}$. The line $k$ with a slope of 2,5 touches $f_p$ in $A$ with $x_A = -1$. Get the function of k algebraically.


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*I might have used wrong terminology, because English is not my native language, I will hopefully clear up doubts on what this problem is by showing what I did.


First off, I got $[f_p(x)]'$. This was EXTREMELY troublesome, and is the main reason why I found this problem challenging, because of all the steps. Can you guys show me the easiest and especially quickest way to get this derivative?
After that, I filled in $-1$ in the derivative and made the derivative equal to $2\dfrac{1}{2}$, this was also troublesome for me, I kept getting wrong answers for a while, again: Can you guys show me the easiest and especially quickest way to solve this?
After you get p it is pretty straightforward. I know this might sound like a weird question, but it basically boils down to: I need quicker and easier ways to do this. I don't want to make careless mistakes, but because the length of these types of question, it ALWAYS happens. Any tips or tricks regarding this topic in general would be much appreciated too. 
Update: A bounty will go to the person with the most clear and concise way of solving this question!

The point $A$ is $(x, \space f_p(x))  = \left(x, \space \dfrac{9\sqrt{x^2+p}}{x^2+2}\right)$ at $x = -1$ this gives $A = \left(-1, \space 3\sqrt{1+p}\right)$.
The line $k$ with a slope of $2.5$ touches $f_p$ in $A$ i.e passes through point $A$ then $k =mx+c$ where $m=2.5$ Then you substitute point $A$ in the equation to get the constant $c$, that is $$3\sqrt{1+p} = 2.5 \times -1 + c \quad \Rightarrow \quad c=  3\sqrt{1+p}+2.5$$
which implies that $$k = 2.5x  +3\sqrt{1+p}+2.5$$ 
If by touch you meant tangent to $f_p$ then the only way to find the value of $p$ is to find the derivative of $f_p$.
Instead of using the quotient rule, one can also apply the product rule on $$f_p(x) = 9(x^2+p)^{1/2}(x^2+2)^{-1}$$ which gives 
$\quad \quad \quad \cfrac {df_p}{dx} = 9 \left[ ((x^2 +p)^{1/2})' \cdot (x^2+2)^{-1} + (x^2 +p)^{1/2} \cdot ((x^2+2)^{-1})'\right ]$ 
$\quad \quad \quad  \quad \quad = 9 \left[ x(x^2 +p)^{-1/2} \cdot (x^2+2)^{-1}\space  -\space 2x (x^2 +p)^{1/2} \cdot (x^2+2)^{-2}\right ]$
Compute $f'_p(-1)$ easily as
$\quad \quad \quad =9 \left[ \left ((-1)((-1)^2 +p)^{-1/2} \cdot ((-1)^2+2)^{-1}\right ) + \left (-2(-1) ((-1)^2 +p)^{1/2} \cdot ((-1)^2+2)^{-2}\right)\right ]$
$\quad \quad \quad =9 \left[ -\cfrac 13(1 +p)^{-1/2}+\cfrac 29 (1 +p)^{1/2} \right ]$
$\quad =\quad-3(1 +p)^{-1/2}+2(1 +p)^{1/2} = 2.5$
Let $X = 1+p$,  then $\cfrac {-3}{\sqrt X}+2\sqrt X = 2.5$.  Solving this equation gives $X = 1+p = 4 \quad \Rightarrow \quad p =3$ and $$k = 2.5x  +8.5$$ 
