I am completely lost on this one, I have no idea what to do at all on it. I know that the derivative of $e^x$ is $e^x$ but that doesn't seem to help here. I am suppose to find the derivative of $y=xe^{-kx}$ I figured I could use the product rule which would be the derivative of e $e^{-kx} (x) + e^{-kx} (1)$ I need to find the derivative of $e^{-kx}$ and for that I got $-e^{-kx}$ I likely made many mistakes because my answer isn't even close.
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This is a combination of product rule and chain rule. To review, the product rule states $$\frac{d f g}{dx} = \frac{df}{dx} g + f \frac{dg}{dx}$$ or in Newton's notation $$(fg)' = f' g + g' f.$$ And as you've identified correctly, $$y' = (x e^{- k x})' = (1) e^{-kx} + x (e^{-kx})',$$ so we just need to determine what $(e^{-kx})'$ is. From previous questions, you know that the chain rule states that $$\frac{df(g(x))}{dx} = \frac{df(g)}{dg} \frac{dg(x)}{dx},$$ or more succinctly $$f'(g(x)) = f'(g) g'(x).$$ So, in $(e^{-kx})'$, we can see that $f(g) = e^g$ and $g(x) = -k x$. You can either memorize how to treat the derivative of a constant times your variable, or you could apply the product rule, again. The product rule gives $$g'(x) = (-k)' x + -k (x)'.$$ So, what are the derivatives of $-k$ and $x$ with respect to $x$? Once you have those values in hand, plug back in. |
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