Solving this problem without Taylor series ( I made an extra assumption) I want to show that the derivative of a (one-time differentiable) function $f(x)$ is given by $$f'(a)= \lim \limits_{h \to 0} \dfrac {1}{h} \displaystyle \int_{-1}^{1} \dfrac{3u}{2} f(a+uh) \ du$$
I used the Taylor's expansion about the point $x=a$ and the conventional $h$ as $uh$ [i.e. $f(x+h)= f(x) + h f'(x) + \cdots$] and got the answer. However, I made an additional assumption that the function $f(x)$ possesses a convergent series about $uh$ and this includes the assumption that the function is differentiable more than once.
Tried applying the Leibniz rule, but could not get to the RHS. How can I prove the result without using the Taylor series?
 A: Hint.
Try the change of variables $y=uh$ to the integral:
$$
\frac32\cdot\frac{1}{h^3}\int_{-h}^{h}yf(a+y)dy
$$
Then you can apply the L'Hopital rule and the fundamental theorem of calculus to get
$$
\lim_{h\to 0 }\frac{h(f(a+h)-f(a-h))}{2h^2}=
\lim_{h\to 0 }\frac{f(a+h)-f(a-h)}{2h}
$$
Now use the differentiability of $f$ at $a$.
A: Let $u=t^{1/3}/h$ so that $$du=\frac{1}{3h}t^{-2/3}\,dt$$ and thus the desired limit is $$\lim_{h\to 0}\frac{1}{2h^3}\int_{-h^3}^{h^3}\frac{f(a+t^{1/3})}{t^{1/3}}\,dt$$ Next use $v=h^3$ to get $$\lim_{v\to 0}\frac{1}{2v}\int_{-v}^{v}\frac{f(a+t^{1/3})-f(a)}{t^{1/3}}\,dt+\frac{f(a) }{2v} \int_{-v} ^{v} t^{-1/3}\,dt$$ Note that the second integral is $0$ and the first integral can be split into integrals over $[-v, 0]$ and $[0,v]$. Thus the limit is split into limit of two terms one of which is $$\lim_{v\to 0}\frac{1}{2v}\int_0^v\frac{f(a+t^{1/3})-f(a)}{t^{1/3}}\,dt$$ and by fundamental theorem of calculus this equals $$\frac{1}{2}\lim_{t\to 0}\frac{f(a+t^{1/3})-f(a)}{t^{1/3}}=\frac{f'(a)}{2}$$ Similarly the other limit also equals $f'(a) /2$ and the desired limit is $f'(a)$.

The above uses the assumption that $f$ is Riemann integrable in some neighborhood of $a$ and derivative $f'(a) $ exists.
