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andidious

  • one year 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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  1. luo
    • one year ago
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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)"?

  2. andidious
    • one year ago
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    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
    • one year ago
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    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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