## 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)

1. luo

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

2. andidious

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.

3. luo

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 ?