How can partial fractions be used for deductions? 
Find partial fractions of the expression,$\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}$
  . Hence deduce that; $\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)}+\frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)}+\frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)}=a+b+c+d-p-q-r-s$ 

My Working
I was able to calculate partial fractions as follows,
$\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}=1+\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)(x-a)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)(x-b)}+..$
But I cannot proceed to deduction part. Highly appreciated if someone can give me a hint to work this out. Thank you!

You already got the correct partial fraction decomposition
$$
\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}
=1+\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)} \cdot \frac{1}{x-a} \\
+  \frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)} \cdot \frac{1}{x-b} \\
+  \frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)} \cdot \frac{1}{x-c} \\
+  \frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)} \cdot \frac{1}{x-d} \, .
$$
With the substitution $x = \frac 1y$ we get
$$ \tag{*}
\frac{(1-py)(1-qy)(1-ry)(1-sy)}{(1-ay)(1-by)(1-cy)(1-dy)}
=1+\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)} \cdot \frac{y}{1-ay} \\
+  \frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)} \cdot \frac{y}{1-by} \\
+  \frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)} \cdot \frac{y}{1-cy} \\
+  \frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)} \cdot \frac{y}{1-dy} \, .
$$
For small $y$ the left-hand side has the development
$$
 \frac{1 - (p+q+r+s)y + O(y^2)}{1-(a+b+c+d)y + O(y^2)}
= (1 - (p+q+r+s)y + O(y^2))(1+(a+b+c+d)y + O(y^2) \\
= 1 + (a+b+c+d-p-q-r-s)y + O(y^2) 
$$
for $y \to 0$. On the right-hand side we have
$$
\frac{y}{1-ay} = y(1+ay + O(y^2)) = y + O(y^2) 
$$
and similarly for the other fractions $\frac{y}{1-by}$, $\frac{y}{1-cy}$
and $\frac{y}{1-dy}$.
Therefore a comparison of the $y$-terms (i.e. the derivates with respect
to $y$ at $y = 0$) in equation $(**)$ gives the desired identity
$$
a+b+c+d-p-q-r-s = \frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)} \\
+\frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)} \\
+\frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)} \, .
$$
