# How to prove that a function is uniquely differentiable

real-analysis

Say that we have function $$f: \mathbb{R} \to \mathbb{R}$$ and it is differentiable for all real values. Also, with $$a,b \in \mathbb{R}$$, the following is also true: $$f'(x)=a$$ for all $$x$$ and $$f(0)=b$$. Then, I am asked to find $$f$$ and prove that it is the only unique differentiable function with these properties.

I think it would be a good idea to use this proposition: If $$f: [a,b] \to \mathbb{R}$$ is differentiable, and for each $$x \in [a,b], f'(x) = 0$$, then $$f$$ is a constant function. So it seems that $$f$$ is $$f(x) = ax + b$$ where its derivative is constant for some $$a$$. But how can I utilize the $$b$$ to conclude that this is unique for $$f$$?

There is a "standard" trick to proving uniqueness in analysis. It consists in showing (or immediately using) that two functions $$f$$ and $$g$$ with certain properties have the same derivative, and thus differ only by an additive constant, and then showing that this constant is $$0$$.
In this specific example it goes like this: Let $$g:\mathbb R\to\mathbb R$$ be another differentiable function with $$g'(x)=a$$ for all $$x\in\mathbb R$$ and $$g(0)=b$$. Then $$(f-g)'(x)=f'(x)-g'(x)=a-a=0,$$ so $$f-g$$ is a constant function. But since $$f(0)-g(0)=b-b=0$$, it must be zero. So we actually have $$f=g$$.
This shows that any function satisfying your desired properties is equal to $$f$$.