differentiating negative powers of e

**kingsolomonsgrave** · October 17th 2012, 11:06 AM

the answer is above and the question is

differentiate $y=xe^-kx$

**tom@ballooncalculus** · October 17th 2012, 11:49 AM

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**Kyo** · October 17th 2012, 07:18 PM

You're going to have to use the product rule, chain rule, and also the derivative of an exponential function.

Assuming you're differentiating with respect to x, and k is a constant: y' = (x)' (e^-kx) + x(e^-kx)' . Think you can get it now?

**Prove It** · October 17th 2012, 08:45 PM

Originally Posted by kingsolomonsgrave

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the answer is above and the question is

differentiate $y=xe^-kx$

$\displaystyle \begin{align*} y &= x\,e^{-k\,x} \\ \frac{d}{dx} \left( y \right) &= \frac{d}{dx} \left( x\, e^{-k\,x} \right) \\ \frac{dy}{dx} &= x \, \frac{d}{dx} \left( e^{-k\, x} \right) + e^{-k\, x}\,\frac{d}{dx}\left( x \right) \\ \frac{dy}{dx} &= x\left( -k \, e^{-k\, x} \right) + e^{-k\, x} \left( 1 \right) \\ \frac{dy}{dx} &= -k\, x \, e^{-k\, x} + e^{-k \, x} \\ \frac{dy}{dx} &= e^{-k\, x} \left( -k\, x + 1 \right) \end{align*}$

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