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andidious Group Title

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)

  • one year ago
  • one year ago

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  1. luo Group Title
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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)"?

    • one year ago
  2. andidious Group Title
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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.

    • one year ago
  3. luo Group Title
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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 ?

    • one year ago
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