Differentiation of \(\rm x^{e^{x}}\) with respect to x is1. \(\rm x^

1.	Differentiation of \(\rm x^{e^{x}}\) with respect to x is1. \(\rm x^{e^{x}}\left[\ln x+{1\over x}\right]\)2. \(\rm x^{e^{x}}e^x\left[\ln x+{1\over x}\right]\)3. \(\rm e^x\left[\ln x+{1\over x}\right]\)4. \(\rm x^{e^{x}}\left[\ln x+{e^x\over x}\right]\)
Answer» Correct Answer - Option 2 : \(\rm x^{e^{x}}e^x\left[\ln x+{1\over x}\right]\) Concept: \(\rm d\over dx\)xn = nxn-1 \(\rm d\over dx\)sin x = cos x \(\rm d\over dx\)cos x = -sin x \(\rm d\over dx\)ex = ex \(\rm d\over dx\)ln x = \(\rm1\over x\) \(\rm d\over dx\)(ax + b) = a \(\rm d\over dx\)tan x = sec2 x \(\rm d\over dx\)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 = e^x (ln x) (∵ log mn = n log m) Differentiating with respect to x, we get \(\rm {1\over y}{dy\over dx} = e^x(\ln x)+e^x\left({1\over x}\right)\) \(\rm {dy\over dx} = y\left[e^x(\ln x)+{e^x\over x}\right]\) \(\boldsymbol{\rm {dy\over dx} = x^{e^{x}}e^x\left[\ln x+{1\over x}\right]}\)

Differentiation of \(\rm x^{e^{x}}\) with respect to x is1. \(\rm x^{e^{x}}\left[\ln x+{1\over x}\right]\)2. \(\rm x^{e^{x}}e^x\left[\ln x+{1\over x}\right]\)3. \(\rm e^x\left[\ln x+{1\over x}\right]\)4. \(\rm x^{e^{x}}\left[\ln x+{e^x\over x}\right]\)

Answer» Correct Answer - Option 2 : \(\rm x^{e^{x}}e^x\left[\ln x+{1\over x}\right]\)

Concept:

Calculation:

Let y = \(\rm x^{e^{x}}\)

Taking log both sides, we get

ln y = ln \(\rm x^{e^{x}}\)

ln y = e^x (ln x) (∵ log mn = n log m)

Differentiating with respect to x, we get

\(\rm {1\over y}{dy\over dx} = e^x(\ln x)+e^x\left({1\over x}\right)\)

\(\rm {dy\over dx} = y\left[e^x(\ln x)+{e^x\over x}\right]\)

\(\boldsymbol{\rm {dy\over dx} = x^{e^{x}}e^x\left[\ln x+{1\over x}\right]}\)