Correct Answer - Option 2 :
\(\rm x^{e^{x}}e^x\left[\ln x+{1\over x}\right]\)
Concept:
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\(\rm d\over dx\)xn = nxn-1
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\(\rm d\over dx\)sin x = cos x
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\(\rm d\over dx\)cos x = -sin x
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\(\rm d\over dx\)ex = ex
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\(\rm d\over dx\)ln x = \(\rm1\over x\)
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\(\rm d\over dx\)(ax + b) = a
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\(\rm d\over dx\)tan x = sec2 x
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\(\rm d\over dx\)f(x)g(x) = f'(x)g(x) + f(x)g'(x)
Calculation:
Let y = \(\rm x^{e^{x}}\)
Taking log both sides, we get
ln y = ln \(\rm x^{e^{x}}\)
ln y = ex (ln x) (∵ log mn = n log m)
Differentiating with respect to x, we get
\(\rm {1\over y}{dy\over dx} = e^x(\ln x)+e^x\left({1\over x}\right)\)
\(\rm {dy\over dx} = y\left[e^x(\ln x)+{e^x\over x}\right]\)
\(\boldsymbol{\rm {dy\over dx} = x^{e^{x}}e^x\left[\ln x+{1\over x}\right]}\)