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How do you differentiate exp-ax2? - Answers
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)`= exalso: (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` = 2xIf 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^2And that is the answer.(e-ax^2)`= -2xae-2ax^2
Bing
How do you differentiate exp-ax2? - Answers
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)`= exalso: (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` = 2xIf 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^2And that is the answer.(e-ax^2)`= -2xae-2ax^2
DuckDuckGo
How do you differentiate exp-ax2? - Answers
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)`= exalso: (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` = 2xIf 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^2And that is the answer.(e-ax^2)`= -2xae-2ax^2
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- og:descriptionI'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)`= exalso: (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` = 2xIf 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^2And that is the answer.(e-ax^2)`= -2xae-2ax^2
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