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andidious
Group Title
If someone can check this:
derive y = x^(e^x)
approach:
lny (dy) = e^x * (lnx) (dx)
1/y (dy) = e^x * 1/x + e^x (lnx) (dx) via product rule
dy/dx = e^x *1/x + e^x(lnx) y
= [e^x *1/x + e^x(lnx)] x^(e^x)
 one year ago
 one year ago
andidious Group Title
If someone can check this: derive y = x^(e^x) approach: lny (dy) = e^x * (lnx) (dx) 1/y (dy) = e^x * 1/x + e^x (lnx) (dx) via product rule dy/dx = e^x *1/x + e^x(lnx) y = [e^x *1/x + e^x(lnx)] x^(e^x)
 one year ago
 one year ago

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luo Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ d }{ dx } y = [e^x *1/x + e^x(lnx)] x ^{e ^{x}}\] the last answer is right ,i think what is the mean of "dy" and "dx" in the notation" lny (dy) = e^x * (lnx) (dx)"?
 one year ago

andidious Group TitleBest ResponseYou've already chosen the best response.0
I suppose it's implicit differentiation. You differentiate each variable first, and later decide which variable you would like to solve in terms of. Someone please correct me if I'm wrong.
 one year ago

luo Group TitleBest ResponseYou've already chosen the best response.1
maybe ,what you want to say is this: \[y = x ^{e ^{x}}\] \[ln(y) = \ln(x ^{e ^{x}} )\] \[\ln(y) = \ln (x ^{e ^{x}}) = e ^{x}.lnx\] derivative both sides of the equation \[(\ln(y))\prime = ( e ^{x} . \ln x)\prime\] \[\frac{ 1 }{ y }. y \prime = e ^{x} .\frac{ 1 }{ x } + e ^{x} . lnx\] \[y \prime = (e ^{x} .\frac{ 1 }{ x} + e^x . lnx) .y\] \[y \prime = (e^x \frac{ 1 }{ x } + e^x .\ln x).x ^{e ^{x}}\] are you ?
 one year ago
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