$\displaystyle Q(x) = \frac{1}{\sqrt{2 \pi}} \int_x^\infty e^{- t^2/2} dt$ $\displaystyle P = Q( \sqrt{2a \cdot u^2} ) \cdot e^{au^2}$ what is the answer if we differentiate P w.r.t. $\displaystyle u^2$ ??? thanks
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Originally Posted by graticcio $\displaystyle Q(x) = \frac{1}{\sqrt{2 \pi}} \int_x^\infty e^{- t^2/2} dt$ $\displaystyle P = Q( \sqrt{2a \cdot u^2} ) \cdot e^{au^2}$ what is the answer if we differentiate P w.r.t. $\displaystyle u^2$ ??? thanks Put $\displaystyle v=u^2$, then: $\displaystyle P(v)=Q( \sqrt{2a v} ) ~e^{av}$ $\displaystyle \frac{d}{dv}P(v)=\frac{\sqrt{2a}}{2\sqrt{v}}~Q'(\s qrt{2a v} ) ~ e^{av}+aP(v) $ RonL
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