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Suppose F(y)=\int _{a}^{y}f(t)\,dt
We know that there exist real numbers x\, and x+\delta \, in [a,b]\,, where a and b are non-equal real numbers.

F(x+\delta )-F(x)=\int _{a}^{{x+\delta }}f(t)\,dt-\int _{a}^{{x}}f(t)\,dt

It can be shown that \int _{a}^{{x+\delta }}f(t)\,dt-\int _{a}^{{x}}f(t)\,dt=\int _{x}^{{x+\delta }}f(t)\,dt

By the mean value theorem: \int _{x}^{{x+\delta }}f(t)\,dt=f(c)\delta for some c in [x,x+\delta ]\,

Hence F(x+\delta )-F(x)=f(c)\delta \,

\Rightarrow {\frac  {F(x+\delta )-F(x)}{\delta }}=f(c)

\Rightarrow \lim _{{\delta \to 0}}{\frac  {F(x+\delta )-F(x)}{\delta }}=\lim _{{\delta \to 0}}f(c)

But as \delta \to 0,\,c\to x by the Sandwich Theorem

Hence \lim _{{\delta \to 0}}{\frac  {F(x+\delta )-F(x)}{\delta }}=f(x)

Therefore, relabelling y as x: \,{\frac  {d}{dx}}\int _{a}^{x}f(t)\,dt=f(x)