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So I have $$ f(x) = \int_{0}^{1-x^2} e^{(t^2)} dt $$

And I'm trying to get the derivative of it. I followed the example at the bottom of http://ltcconline.net/greenl/courses/105/antiderivatives/secfund.htm to do this.

So first made $$u = 1-x^2$$ and $$y = \int_{0}^{u}e^{(t^2)} dt$$

then used the chain rule $$\frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx} = e^{u^2} * (-2x)$$ Then subbed u for 1-x^2 in, giving me $$= -2e^{(1-x^2)^2}x$$

But when I plug this in WolframAlpha, I get $$\frac{d}{dx}(\int_{0}^{1-x^2} e^{(t^2)} dt) = -2 e^{(x^2-1)^2} x$$

In the exponent of e, mine is 1-x^2, but Wolfram gives me x^2-1?

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1 Answer 1

up vote 1 down vote accepted

note that $$ y^2 = (-y)^2 \Rightarrow (x^2-1)^2 = (1-x^2)^2 $$

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Where in my answer did I go wrong? Are both answers correct? –  Rickkwa Sep 14 '13 at 16:24
Both answers are the same, that's what what's up is trying to convey to you! –  DonAntonio Sep 14 '13 at 16:35
Thank you very much –  Rickkwa Sep 14 '13 at 16:45

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