# 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 $$a+b-c+d$$.

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