# first partial derivative of $F(x,y)=\int_a^x f(t,y)dt$

According to the fundamental theorem of calculus, the first partial derivative is f(x,y).

I'm wondering why I can't apply L'Hopital's rule in the following reasoning:

$$\lim_{h\to0}\frac{\int_a^{x+h}f(t,y)dt - \int_a^x f(t,y)dt}{h}=\lim_{h\to 0}\frac{f(x+h,y)-f(x,y)}{1}=0$$

While the correct argument should be:

\begin{align*}\lim_{h\to0}\frac{\int_a^{x+h}f(t,y)dt - \int_a^x f(t,y)dt}{h}&=\lim_{h\to 0}\frac{\int_a^{x+h}f(t,y)dt+\int_x^a f(t,y)dt}{h}=\lim_{h\to 0}\frac{\int_x^{x+h}f(t,y)dt}{h}\\&=\lim_{h\to 0}\frac{f(c,y)h}{h}=\lim_{h\to 0}f(c,y)=f(x,y)\end{align*} where $c\in [x,x+h].$

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Check your first equation again... – t.b. Jun 14 '11 at 18:34
The title has a $dy$ when you want a $dt$. – Thomas Andrews Jun 14 '11 at 18:40

$$\lim_{h\to0}\frac{\int_a^{x+h}f(t,y)dt - \int_a^x f(t,y)dt}{h}=\lim_{h\to 0}\frac{f(x+h,y)}{1}=f(x,y) \; .$$
Since the second term is independent of $h$.
Yes, your limit is as $h\rightarrow 0$, and you differentiate the denominator against $h$, so the numerator should be differentiated against $h$, too. – Thomas Andrews Jun 14 '11 at 18:40