Truncation Error in differentiation I dont understand how truncation error ie, the error caused because the definition of h tends to zero, but we take the value(during differentiation) as a non zero really small number; gets reduced because of taking centre difference. 
That is, take $\frac{f(x+h)−f(x−h)}{2h}$ instead of $\frac{f(x+h)−f(x)}{h}$. How does this reduce the truncation error ?

By taking symmetric steps around the centre, we're able to cancel out the first-order term to achieve a second-order formula. Indeed, consider their Taylor expansions:
\begin{align*}
f(x + h) &= f(x) + f'(x)h + \frac{1}{2}f''(x)h^2 + \frac{1}{6}f'''(x)h^3 + \cdots \\
f(x - h) &= f(x) - f'(x)h + \frac{1}{2}f''(x)h^2 - \frac{1}{6}f'''(x)h^3 + \cdots 
\end{align*}
Subtracting the two equations and dividing through by $2h$, we obtain:
$$
\frac{f(x + h) - f(x - h)}{2h}
= f'(x) + \frac{1}{6}f'''(x)h^2 + \cdots \\
\implies \boxed{f'(x) = \frac{f(x + h) - f(x - h)}{2h} \color{red}{+ O(h^2)}}
$$
On the other hand, if we took the first equation, subtracted $f(x)$, then divided by $h$, we would obtain:
$$
\frac{f(x + h) - f(x)}{h}
= f'(x) + \frac{1}{2}f''(x)h + \frac{1}{6}f'''(x)h^2 + \cdots \\
\implies \boxed{f'(x) = \frac{f(x + h) - f(x)}{h} \color{red}{+ O(h)}}
$$
