So I have the function $$ e^{-2x} $$ and if I derive this I thought that I should get $$ -2xe^{-2x} $$ But the $x$ disappears, why? Is it an inner derivative and because of that, I also have to differentiate the expression $-2x$ when I put it in front of $e$? If that is the case, then $x$ would be 1 and -2 is the only character left.. Am I right?
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4$(-2x)'=-2$, no? – David Mitra Jan 25 '14 at 13:13
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2"differentiate (obtain derivative)", not "derive (prove)". – user21820 Jan 25 '14 at 13:16
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@theva Yes, what you say in your last paragraph is correct. – Arthur Jan 25 '14 at 13:22
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Yes, you are correct in your thoughts in the last paragraph.
We use the chain rule:
Let $u = -2x.\;$ Then $\;\dfrac {du}{dx} = -2.$
This gives us $$\frac{d}{dx}(e^{-2x}) = \frac d{dx}(e^u) = e^u\left(\dfrac{du}{dx}\right) = e^{-2x}(-2) = -2e^{-2x}$$
amWhy
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Go back to the basic definition of derivative. You have this:
\begin{align} \frac{d}{dx}\left(e^{-2x}\right) &= \lim_{h\to0} \frac{e^{-2(x+h)}-e^{-2x}}{h}\\ &=\lim_{h\to0} \frac{e^{-2x}(e^{-2h}-1)}{h}\\ &=e^{-2x}\lim_{h\to0} \frac{e^{-2h}-1}{h} \end{align} As you can see, there is no $x$ outside of the $e^{-2x}$. And you will find that the remaining limit is equal to $-2$.
Glen O
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