# Thread: Differentiating an integral-defined function

1. ## Differentiating an integral-defined function

(I think) It is well known that for an integral-defined function, differentiating them with the upper limit(i.e. 'x') takes the integral away. (*) i.e.
$\frac{d}{dx}\int_{0}^{x}\ f(t) \ dt = f(x)$

The problem is, I came across a weird question today:
determine $\frac{d}{dx}\int_{0}^{x}\frac{e^{t}}{e^t+e^{2x}} \ dt$

Well, integrating $\frac{e^{t}}{e^t+e^{2x}}$ first and differentiating yields the (probably) correct answer
$\frac{1+2e^x-{e^{2x}}}{(e^x+1)(e^{2x}+1)}$

But when i apply the rule (*) 'differentiation of integral-defined function', it gives a completely different result
$\frac{e^x}{e^x+e^{2x}}$

Can anybody tell why?

2. ## Re: Differentiating an integral-defined function

Oh, and as far as i can think, the constant $e^{2x}$ seems to be the problem. the rule always holds when the integrated function does not contain terms related to the differentiation.(and only consists variables of the integration)

But i cannot figure out why the constant mess with the rule

3. ## Re: Differentiating an integral-defined function

If I were going to evaluate this derivative, I would use the substitution:

$u=e^t+e^{2x}\,\therefore\,du=e^t\,dt$

and now we have:

$\frac{d}{dx}\int_{1+e^{2x}}^{e^x+e^{2x}}\frac{1}{u }\,du=\frac{e^x+2e^{2x}}{e^x+e^{2x}}-\frac{2e^{2x}}{1+e^{2x}}=$

$\frac{1+2e^{x}}{1+e^{x}}-\frac{2e^{2x}}{1+e^{2x}}=\frac{1+2e^x-e^{2x}}{\left(1+e^{x} \right)\left(1+e^{2x} \right)}$

4. ## Re: Differentiating an integral-defined function

It's because, for your rule, the contribution from the lower limit is independent of x and so gets differentiated out.
For your actual problem, the contribution from the lower limit is a function of x.
For your actual problem, the rule does not apply.

5. ## Re: Differentiating an integral-defined function

Ohh.. indeed! Thanks!