to evaluate using L'hopital's rule Can anyone please guide me how to  use L'Hop rule to evaluate this 
$$
f(x)=\lim_{t\to 0} \frac{1}{2t} \int_{x-t}^{x+t} s f'(s) ds
$$
 A: Think of the integral $$\int_{x-t}^{x+t}sf'(s)ds = \int_{x-t}^x sf'(s)ds + \int_x^{x+t} sf'(s)ds$$
and now use the Fundamental Theorem of Calculus to differentiate each of these with respect to $t$.
If you're not sure how to do the first one, remember that $\displaystyle\int_{x-t}^x sf'(s)ds=-\displaystyle\int_x^{x-t} sf'(s)ds$.
A: $$f(x)=\lim_{t\to 0} \frac{1}{2t} \int_{x-t}^{x+t} s f'(s) ds$$
Using L'Hospital rule:
$$
=\lim_{t\to 0}\frac{\frac{d}{dt}\left(\int_{x-t}^{x+t} s f'(s) ds\right)}{\frac{d}{dt}(2t)}$$
Using Extended-Leibniz Rule:$$
f(x)=\lim_{t\to 0}\frac{\int_{x-t}^{x+t}\frac{\partial( s f'(s))}{\partial t} ds+(x+t)f'(x+t)\frac{\partial(x+t)}{\partial t}-(x-t)f'(x-t)\frac{\partial(x-t)}{\partial t}}{2}\\f(x)=\lim_{t\to0}\frac{(x+t)f'(x+t)+(x-t)f'(x-t)}{2}\\f(x)=2f'(x)$$
Or:
$$\frac{f'(x)}{f(x)}=\frac12\implies \ln f(x)=\frac{x}2+c\implies f(x)=ke^{x/2}$$

For the Extended-Leibniz rule:
$$\frac{d}{d\theta}\left(\int_{a(\theta)}^{b(\theta)}f(x,\theta)dx\right)=\int_{a(\theta)}^{b(\theta)}\frac{\partial f(x,\theta)}{\partial\theta}dx+f(b(\theta),\theta).\frac{\partial b(\theta)}{\partial\theta}-f(a(\theta),\theta).\frac{\partial a(\theta)}{\partial\theta}$$
