# Calculuszation

**Calculus**Level 4

Let \(g(x)\) denotes the first derivative of \(f(x)\) such that \(f(1)=2\) and \(g(1)=144\). If the \(99th\) derivative of \(g(x)\) is in the form \(\frac{a!}{b!}x^{c}+d\), where \(a,b,c\) and \(d\) are non-negative integers, find the value of \(a+b-c+d\).