derivative $\displaystyle \int_{0}^{\sqrt{x}} e^{-t^2} $
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Originally Posted by peteryellow derivative $\displaystyle \int_{0}^{\sqrt{x}} e^{-t^2} $ this is the second question you've asked regarding this topic. note that u is a function of x ... $\displaystyle \frac{d}{dx} \left[ \int_a^u f(t) dt\right] = f(u) \cdot \frac{du}{dx}$
Last edited by mr fantastic; Aug 30th 2009 at 08:12 PM.
how is this true? $\displaystyle \frac{d}{dx} \left[ \int_a^u f(t) dt\right] = f(u) \cdot \frac{du}{dx} $?
$\displaystyle \frac{d}{dx}\int_{0}^{\sqrt{x}} e^{-t^2} = e^{-(\sqrt{x})^2} \frac{d}{dx}(\sqrt{x} - 0) = ? $ Can you finish it?
Originally Posted by peteryellow how is this true? $\displaystyle \frac{d}{dx} \left[ \int_a^u f(t) dt\right] = f(u) \cdot \frac{du}{dx} $? It follows directly from the chain rule.
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