I was wondering if I can apply the FTOC: $\frac { d }{ dx } \left( \int _{ a }^{ x }{ f(t) } dt \right) =f(x)$ to an implicit function with the variable being differentiated implicit in the function.

For example: $\frac { d }{ dx } \left( \int _{ a }^{ x }{ xf(t) } dt \right)$ or any function in the form: $\frac { d }{ dx } \left( \int _{ a }^{ x }{ f(x,t)f(t) } dt \right) $ and if so, what would we get.

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    $\begingroup$ Note $\int_a^x x f(t) \mathop{dt} = x \int_a^x f(t)\mathop{dt}$ $\endgroup$
    – angryavian
    Mar 6, 2016 at 9:10
  • $\begingroup$ Would this be the same if we have say: $\frac { d }{ dx } \left( \int _{ a }^{ x }{ f(x,t)f(t) } dt \right) $ such as $\frac { d }{ dx } \left( \int _{ a }^{ x }{ \sqrt { 1+xt } f(t) } dt \right) $ or would we have to deal with it otherwise? $\endgroup$ Mar 6, 2016 at 9:14
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    $\begingroup$ May be simply $\frac{d}{dx}(\int_a^x f(x,t)dt)$. $\endgroup$
    – velut luna
    Mar 6, 2016 at 9:19
  • $\begingroup$ "wander" $\ne$ "wonder". $\endgroup$
    – user21820
    Mar 6, 2016 at 9:27
  • $\begingroup$ Sorry, English is not my first language. Edited. $\endgroup$ Mar 6, 2016 at 9:28

2 Answers 2


By definition $$\frac{d}{dx}\int_a^xf(x,t)dt=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[\int_a^{x+\Delta x}f(x+\Delta x,t)dt-\int_a^x f(x,t)dt\right]$$ $$=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[\int_a^{x+\Delta x}\left(f(x,t)+\frac{\partial f}{\partial x}\Delta x\right)dt-\int_a^x f(x,t)dt\right]$$ $$=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[\int_a^{x+\Delta x}f(x,t)dt-\int_a^x f(x,t)dt + \int_a^{x+\Delta x}\frac{\partial f}{\partial x}\Delta x dt\right]$$ $$=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[\int_x^{x+\Delta x}f(x,t)dt + \int_a^{x+\Delta x}\frac{\partial f}{\partial x}\Delta x dt\right]$$ $$=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\left[f(x,x)\Delta x + \Delta x \int_a^{x+\Delta x}\frac{\partial f}{\partial x}dt\right]$$ $$=f(x,x)+\lim_{\Delta x\to 0} \int_a^{x+\Delta x}\frac{\partial f}{\partial x}dt$$ $$=f(x,x)+\int_a^x \frac{\partial}{\partial x}f(x,t)dt$$

In conclusion, you will only get the first term if you apply the FTOC blindly. Because the integrand contains $x$, you will have the second extra term.

  • $\begingroup$ Where did $\int_x^{x+\Delta x}f(x,t)dt=f(x,x)\Delta$ come from? $\endgroup$ Jan 25, 2018 at 10:37

In mathematics any question of the form:

Can we do X?

always has the answer:

If you have proven that you can.

In particular if you ask:

Can you apply theorem X to Y?

the answer is just:

As long as Y satisfies the conditions required by theorem X.

In your case, the fundamental theorem of calculus does not apply when the integrand involves the limit of integration.

You may be wondering whether the theorem can somehow be extended to cover that case. The simple answer is that it cannot, and nothing more can be said unless we know exactly what kind of integrand it is.

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    $\begingroup$ First, thanks for replying, Second, if I may rephrase my question, we know by the FTOC that if g is a function of x: $g(x)=\int _{ a }^{ x }{ f(t)dt } $ , then $g'(x)=f(x)$ , so my question would be if instead $h(x)=\int _{ a }^{ x }{ f(x,t)dt } $, would $h'(x)=f(x,x)$ similarly as we regard x as a constant inside the integral?. $\endgroup$ Mar 6, 2016 at 9:44
  • $\begingroup$ So for example if $h(x)=\int _{ a }^{ x }{ \sqrt { 1+xt } f(t)dt } $, is it true that we say $h'(x)=\sqrt { 1+x^{ 2 } } f(x)$? $\endgroup$ Mar 6, 2016 at 9:46
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    $\begingroup$ I am sorry if I irritated you in some way, and thank you I think I got it. The simplest example if $\int _{ a }^{ x }{ xf(t)dt } $, and since x is irrelevant of t we can take it out to get $x\int _{ a }^{ x }{ f(t)dt } $. If I differentiate wrt x, I get $\frac { d }{ dx } \left( x\int _{ a }^{ x }{ f(t)dt } \right) =xf(x)+\int _{ a }^{ x }{ f(t) } dt$ which is not the same as xf(x) as long as $x\neq a$. $\endgroup$ Mar 6, 2016 at 9:58
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    $\begingroup$ @AspiringMat: No you didn't irritate me, but I am earnestly telling you to always check every tiny detail of the conditions required for a theorem. It is often overlooked by many students who just want the answer and not understand the logical reasoning completely. As for the simplest example, I would give $\int_0^x x\ dt = [xt]_0^x = x^2$ which when differentiate with respect to $x$ gives $2x$, not $x$. $\endgroup$
    – user21820
    Mar 6, 2016 at 10:03
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    $\begingroup$ @AspiringMat: As for your example, your last phrase is not correct. It may be that $x \ne a$ but $\int_a^x f(t)\ dt = 0$. However, you got the idea, that if that integral is not zero then it gives a desired counter-example. $\endgroup$
    – user21820
    Mar 6, 2016 at 10:05

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