differentiating the generating function of the Legendre equation I need to differentiate the generating function 
$$G(x,t)=\sum a_n(x)t^n$$ w.r.t. x in order to solve 
$$\tfrac{d}{dx}[(1-x^2)\tfrac{dG}{dx}]+\tfrac{d}{dt}[t^2\tfrac{dG}{dt}]$$. 
But I don't understand how this can be done ..... any suggestions ?
context:I know that not giving an attempt is frowned upon in this community. I almost always do. In this particular instance this is part of a much larger question which I have in fact given a big attempt at but it is too long to type out. I do'nt want the question done for me I just want to know how to differentiate G wrt x
Note: these were supposed to be partial derivatives ( i didn't know the symbol)
Note : $G(x,t)=(1-2xt+t^2)^{-1/2}$
 A: While taking partial derivatives, you treat the other variable as if it were a constant :
$\frac{\partial{G}}{\partial{x}} = \frac{\partial{{(1 - 2xt + t^2)}^{-1/2}}}{\partial{x}}$
$= {-1 \over 2}\cdot{(1-2xt + t^2)}^{-3/2}\cdot \frac{\partial{(1-2xt+t^2)}}{\partial{x}}$ (Using the chain rule)
$= {-1 \over 2}\cdot{(1-2xt + t^2)}^{-3/2}\cdot(-2)$
$= {(1-2xt + t^2)}^{-3/2}$

UPDATE -
For the series version, simply use the same procedure (all variables except $x$ are to be treated as constants , mentally replace variable $t$ by constant $a$; $x \mapsto t$ ). Here you will get something along the lines of :
$\frac{\partial}{\partial{x}} { \{ \Sigma (a_n(x) \cdot (t)^n) \} } = \Sigma \{ \frac{\partial{ (a_n(x) \cdot (t)^n) }}{\partial{x}} \}$
$= \Sigma \{ (t)^n \cdot \frac{\partial{a_n(x)}}{\partial{x}} \}$

However, I do not think that is what you should be looking for.
What you need is :
$$  \frac{\partial{G}}{\partial{x}} = \frac{1}{{(1-2xt + t^2)}^{3/2}}$$
and
$$ \frac{\partial{G}}{\partial{t}} = \frac{(x-t)}{{(1-2xt + t^2)}^{3/2}} $$
Be sure to accept if that solves your problem ;) (@exodius)
