Verified answer

y = x^(ln x)

take logs

ln y = ln x * ln x

ln (y) = ( ln x)^2

differentiate

y ' /y = 2 ln(x) /x

y ' = (y/x) ln(x^2)

y ' = (1/x) x^(ln x) ln(x^2)

y ' = x^[ln(x) - 1 ] ln(x^2)

b)

xy + e^(x + y) = 0

xy + e^x * e^y = 0

differentiating implicitly

xy' + y + e^x*( y' e^y ) + e^y *e^x = 0

y ' (x + e^x e^y ) + y + e^x e^y = 0

y ' (x + e^(x+y)) = - (y + e^(x+y))

y ' = - [ y + e^(x+y) ] / [ x + e^(x+y) ]

1) y = x^(lnx)

Take ln both sides

ln y = lnx * lnx

ln y = (lnx)^2

Take derivative both sides

y' / y = 2(lnx) * (1/x)

y' = y [(2lnx)/x]

y' = x^(lnx) [(2lnx)/x]

2) xy + e^(x + y) = 0

Take derivative both sides

(1)y + x*y' + e^(x+y) * (1+y') = 0

y + xy' + e^(x+y) + e^(x+y) y' = 0

xy' + e^(x+y)y' = - (y+e^(x+y))

y' (x + e^(x+y)) = - (y + e^(x+y))

y' = -(y + e^(x+y)) / (x+e^(x+y))

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