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anonymous
 one year ago
How do you apply the power rule when E is raised by an power to a power?
f[x] = e^(x^2)
f'[x] = ???
anonymous
 one year ago
How do you apply the power rule when E is raised by an power to a power? f[x] = e^(x^2) f'[x] = ???

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TheSmartOne
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino QH help :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0derivative or integration?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1You don;t use the power rule here. Here you need the derivative of e^u, where u is a function in x.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i see, derivative, chain and then power

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0trying to get to derivative.. yes

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1\(\dfrac{d}{dx} e^u = e^u \dfrac{d}{dx}u\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here we have to apply the derivation of composed functions

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Oh, I see. The power rule does come in to differentiate the exponent, x^2.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so its like 2 functions?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2noe we can write your function as follows: \[f\left( x \right) = {e^{g\left( x \right)}},\quad g\left( x \right) = {x^2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so the requested first derivative is given by the subsequent formula: \[f'\left( x \right) = \frac{{df}}{{dg}} \times \frac{{dg}}{{dx}} = {e^{g\left( x \right)}} \times 2x = 2x{e^{{x^2}}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wicked .. thank you now I just need to get my head around that.. :) ... first day solving derivative problems using rules. .. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0makes sense though :)
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