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lgbasallote
Group Title
LGBADERIVATIVE:
\[\huge \frac{\text d}{\text dx} x^{e^{x^2}}\]
 2 years ago
 2 years ago
lgbasallote Group Title
LGBADERIVATIVE: \[\huge \frac{\text d}{\text dx} x^{e^{x^2}}\]
 2 years ago
 2 years ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.0
rewrite as \[e^{e^{x^2}\ln(x)}\] and use the chain rule
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
do you mean \[\ln y = e^{x^2} \ln x?\]
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.1
\[\huge x^{e^{x^2}}*lnx* e^{x^2}*2x\]
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
uhh how?
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.1
\[\huge y=a^{f(x)}\] \[\huge y'=a^{f(x)}*f'(x)*lna\]
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
oh i see
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
yeah that makes sense thanks
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.1
I made a mistake..
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
x is not constant that formula is for constants
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.1
\[\huge x^{e^{x^2}}*\left( e^{x^2}*2x*lnx+\frac{e^{x2}}{x}\right)\]
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.1
@mukushla you're right, but still we need that formula..
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.1
dw:1341688029055:dw
 2 years ago
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