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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)

MIT 18.01 Single Variable Calculus (OCW)
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\[\frac{ d }{ dx } y = [e^x *1/x + e^x(lnx)] x ^{e ^{x}}\] the last answer is right ,i think what is the mean of "dy" and "dx" in the notation" lny (dy) = e^x * (lnx) (dx)"?
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.
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 ?

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