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Let's say we have:

$$y = \int_0^x e^{t^2} dt$$

and we want $dy/dx$.

Do we consider $t$ as being $x$?

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err i meant dy/dx –  Gladstone Asder May 5 '13 at 23:01
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Refer to Fundamental Theorem of Calculus. –  NasuSama May 5 '13 at 23:08
    
Ditto what Gladstone said. Got a calc book? Look it up. And the answer to your last question is "no". –  Stefan Smith May 5 '13 at 23:26
    
Would it be different if 0 was 2x instead? –  Gladstone Asder May 5 '13 at 23:39
    
@GladstoneAsder: Yes of course since you get : $$\lim_{h\to 0}\frac 1h\left[\int_{2x+2h}^{x+h} e^{t^2} dt-\int_{2x}^{x} e^{t^2} dt\right]=\lim_{h\to 0}\frac 1h\left[\int_x^{x+h} e^{t^2} dt-\int_{2x}^{2x+2h} e^{t^2} dt\right]\\=e^{x^2}-2\;e^{(2x)^2}$$ (you have to subtract at the bottom, $t\mapsto 2x$ and the interval is doubled!). –  Raymond Manzoni May 6 '13 at 7:48
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5 Answers 5

up vote 3 down vote accepted

HINT: Let $$y = \int^x_a f(t) dt$$ if we know the anti-derivative of $f$ is $F$, then by fundamental theorem of calculus $$ y = F(x) - F(a) $$ Hence $$ \frac{dy}{dx} = \frac{d}{dx}F(x) - \frac{d}{dx}F(a) = \frac{d}{dx}F(x) = f(x). $$

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Shortly yes. Longer derivation :

\begin{align} y'(x):=\lim_{h\to 0}\frac{y(x+h)-y(x)}h&=\lim_{h\to 0}\frac 1h\left[\int_0^{x+h} e^{t^2} dt-\int_0^{x} e^{t^2} dt\right]\\ &=\lim_{h\to 0}\frac 1h\int_x^{x+h} e^{t^2} dt\\ &=e^{x^2}\\ \end{align}

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$\dfrac{dy}{dx} = e^{x^2}$ by a direct use of FTC. Note that $t$ is a dummy variable.

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In general we have by chain rule $$\frac{d}{dx}\left( \int_{v(x)}^{u(x)} f(t) dt\right)=f\left(u(x)\right)\times u'(x)-f\left(v(x)\right)\times v'(x)$$

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A related problem. $$ y(x+h)-y(x) = \int_0^{x+h} e^{t^2} dt - \int_0^{x} e^{t^2} dt $$

$$ = \int_0^{x} e^{t^2} dt + \int_{x}^{x+h} e^{t^2} dt - \int_0^{x} e^{t^2} dt $$

$$ \implies \frac{y(x+h)-y(x)}{h}= \frac{1}{h}\int_{x}^{x+h} e^{t^2} dt $$

Take the limit of the above as $h\to 0$ and see what you get.

Added: For the sake of completion. To see that the limit is $e^{x^2}$, see this

$$ \Big| \frac{y(x+h)-y(x)}{h}-e^{x^2} \Big|= \Big|\frac{1}{h}\int_{x}^{x+h} e^{t^2} dt-\frac{1}{h}\int_{x}^{x+h} e^{x^2} dt \Big| $$

$$ =\Big| \frac{1}{h}\int_{x}^{x+h} \left(e^{t^2}-e^{x^2}\right) dt \Big|\leq \frac{1}{h}\int_{x}^{x+h}\Big| e^{t^2}-e^{x^2} \Big|dt $$

$$ < \epsilon \frac{1}{h}\int_{x}^{x+h} dt = \epsilon. $$

Note that we used the continuity of the function on the interval $[x,x+h]$, that is

$$ |h|<\delta \implies \Big| e^{t^2}-e^{x^2} \Big|< \epsilon . $$

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