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I'm not sure if you are asking about: e-ax2 or e-ax^2. Use the carets=^ in the future to raise something to a power.
For e-ax2:
You need to remember that: (ex)`= ex
also: (eax)`= aeax, assuming a is a constant and not a function of x.
Justrecognizeany constant drops down to the front.
Therefore: (e-ax2)`=-2ae-ax2, treat the negative just as a constant or (-1)
For e-ax^2: this one is a little bit trickier:
We can just use u-substitution.
Let u = x2.
u` = 2x
If we substitute the u for x^2, we have: e-au , and we know the deriviate of that is -ae-2au, but we need to multiply by u` as well to get:
(-ae-2au)(2x) = -2xae-2au, then we can just plug u back in.
u = x2, so:
-2xae-2au = -2xae-2ax^2
And that is the answer.
(e-ax^2)`= -2xae-2ax^2
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