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andidious
 2 years ago
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)
andidious
 2 years ago
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)

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luo
 2 years ago
Best 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)"?

andidious
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.

luo
 2 years ago
Best ResponseYou've already chosen the best response.1maybe ,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 ?
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