Thread: differentiating negative powers of e

1. differentiating negative powers of e

the answer is above and the question is

differentiate $y=xe^-kx$

2. Re: differentiating negative powers of e

Just in case a picture helps...

... where (key in spoiler) ...

Spoiler:

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

But this is wrapped inside the legs-uncrossed version of...

... the product rule, where, again, straight continuous lines are differentiating downwards with respect to x.

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3. Re: differentiating negative powers of e

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?

4. Re: differentiating negative powers of e

Originally Posted by kingsolomonsgrave

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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differentiating negative powers

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