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According to the fundamental theorem of calculus, if $f$ is continuous and $F$ is defined as $F(x)=\int_a^x f(t) dt$ then $F'=f$. But what happens if $x$ appears inside the integral? I'm trying to find the derivattive of

$$f(x)=\int_0^x \frac{e^{xy}}{y}dy$$

I read about the Libniz integral rule, but when I try to use it I get

$$\frac{df(x)}{x}=\int_0^x e^{xy} dy + \frac{e^{x^2}}{x}-\frac{1}{0}\cdot0$$

Is this because $\frac{e^{xy}}{y}$ is not defined when $y=0$? Also, is the Leibniz rule the only way to solve problems like this one?

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Do you mean $\frac{df(x)}{dx}$? –  Calvin Lin Jan 4 '13 at 2:12
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up vote 1 down vote accepted

The main hypotheses for applying the Leibniz integral rule is that the derivative of $e^{xy}/y$ in $x$ must exist and be continuous on a rectangle whose base contains the domain of integration. Clearly $e^{xy}/y$ cannot be extended to continuously to any point at which $y=0$ due to the division by 0, so it certainly cannot be differentiable. Thus it fails the main hypothesis.

If you wanted to calculate $f'(x)$ in this case you would have to do it by hand at $x=0$, but the Leibniz formula you derived should work fine for $x\neq 0$ as a $\int_0^{x+h}-\int_0^{x} = \int_x^{x+h}$ stays away from the bad point when $x\neq 0$ (so long as you first prove these integrals exist).

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