Question

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)

Answer 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)"?

Answer 2

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.

Answer 3

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 ?