# Differentiate under integral sign

Let $f$ is function with continuous derivatives, such that $f(\sqrt{2})=2$ and for any real numbers $x$, $$f(x)=\lim_{t \to 0}{\dfrac{1}{2t}\int_{x-t}^{x+t}sf'(s)ds}.$$

I have tried but not able able to get the correct solution by Leibniz's Rule but got correct answer by " Fundamental Theorem Of Calculus ".Thanks for help in advance.

• Please restate the first sentence. It doesn't make much sense as it stands. – zhw. Aug 18 '17 at 15:54
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Note that by the Fundamental Theorem Of Calculus, \begin{align*} \lim_{t \to 0}\dfrac{1}{2t}\int_{x-t}^{x+t}sf'(s)ds&= \dfrac{1}{2}\lim_{t \to 0}\dfrac{1}{t}\int_{x}^{x+t}sf'(s)ds +\dfrac{1}{2}\lim_{t \to 0}\dfrac{1}{(-t)}\int_{x}^{x-t}sf'(s)ds\\ &=\dfrac{1}{2}xf'(x)+\dfrac{1}{2}xf'(x)=xf'(x). \end{align*} Hence $f$ solves the differential equation $f(x)=xf'(x)$ which implies $f(x)=cx$ with $c\in\mathbb{R}$. By using the initial condition $f(\sqrt{2})=2$ we obtain that $f(x)=\sqrt{2}x$.
• The last equality in the set of equations should just be $f(x)$, no? – Harry Aug 18 '17 at 17:29
Hint: Use Hospital rule and with Leibniz integral rule $$f(x)=\lim_{t \to 0}\dfrac{\int_{x-t}^{x+t}sf'(s)ds}{2t}=\dfrac{xf'(x)+xf'(x)}{2}=xf'(x)$$