y= x^(e^x)

To differentiate, first we will apply the natural logarithm to both sides:

==> lny = ln [x^(e^x)]

We know that: ln a^b = b*ln a

==> lny = (e^x) * ln x

Now we will differentiate both sides:

==> (lny)' = [e^x)*lnx]'

To differentiate e^x * ln x we will use the product rule:

[(e^x)*lnx]' = (e^x)'*lnx + (e^x)*(lnx)'

= (e^x)lnx + e^x *1/x

==> (1/y) y' = (e^x)*lnx + e^x(1/x)

==> (y'/y) =( e^x)( lnx/x)

**==> y' = y*(e^x)*lnx /x **

To differentiate y = x^(e^x)).

Solution:

y = x^(e^x).

We take logarithm of both sides:

ln y = e^x lnx.

Now we differentiate both sides with respect to x:

(lny)' = (e^x lnx)'

(1/y)dy/dx = (e^x)'lnx +e^x (lnx)'

(1/y)dy/dx = e^xlnx+ e^x/lx .

dy/dx = y {e^xlnx+e^x /x).

dy/dx = y e^x{lnx+1/x).

dy/dx ={ x^(e^x)} e^x{lnx+1/x}.

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