Hi I need help with this exercise:
Let $f:\;[a,b]\rightarrow\mathbb{R}$ be a differentiable function such that $f'$ is continuous on $(a,b)$. Prove for all $\epsilon>0$ there exist $\delta>0$ such that $$\left|\frac{f(t)-f(x)}{t-x}-f'(x)\right|<\epsilon$$ if $0<|t-x|<\delta$, $\ $ $a<x,t\leq b$.
Please if you can, get me some ideas for do this. I know I need do it for definition, but I don't know how to start.
Let's assume that $f'$ continuous on all of $[a,b]$. Then $f'$ is uniformly continuous on $[a,b]$.
Given an $\epsilon>0$ there is a $\delta>0$ such that $|f'(y)-f'(x)|<\epsilon$ whenever $x$, $y\in[a,b]$ and $|y-x|<\delta$. Assume that $t$, $x\in[a,b]$ with $0<|t-x|<\delta$. Then there is a $y$ between $t$ and $x$ such that $${f(t)-f(x)\over t-x}=f'(y)\ .$$ Since $|y-x|\leq |t-x|<\delta$ we can be sure that $|f'(y)-f'(x)|<\epsilon$, and the claim follows.
But one can say more:
Theorem. There is a continuous function $\phi:\>[a,b]^2\to{\mathbb R}$ such that $$\phi(x,y)={f(y)-f(x)\over y-x}\quad(y\ne x),\qquad\phi(x,x)=f'(x)\ .$$ Proof. Write $$f(y)-f(x)=\int_x^y f'(t)\>dt=(y-x)\int_0^1f'\bigl((1-\tau)x+\tau y\bigr)\>d\tau\ .$$ It follows that $\phi(x,y):=\int_0^1f'\bigl((1-\tau)x+\tau y\bigr)\>d\tau$ does the job.