differentiate xx+x1/x

Ans. 

y = {tex}x^x + x^{1\over x}{/tex}

Let {tex}y = u + v {/tex}

where {tex}u = x^x \ \ \ and \ \ v = x^{1\over x}{/tex}

then 

{tex}{dy\over dx} = {du\over dx} + {dv\over dx}{/tex}  ............... (1)

Now, 
{tex}u = x^x {/tex}
Taking log both sides, 

{tex}log \ u = log\ x^x{/tex}

{tex}=> log \ u = xlog\ x{/tex}

Differentiate w.r.to x, we get 

{tex}=> {1\over u}{du\over dx} = \left (x \times {1\over x} \right ) + log \ x{/tex}

{tex}=> {1\over u} {du\over dx}= 1 + log \ x{/tex}

{tex}=> {du\over dx}= x^x(1 + log \ x){/tex} ........... (2)

Now, {tex}v= x^{1\over x}{/tex}

Taking log both sides, 

{tex}=> log \ v = log \ x^{1\over x}{/tex}
{tex}=> log \ v = {1\over x}log \ x{/tex}

Differentiate w.r.to x 

{tex}=> {1\over v} {dv\over dx} = \left ( {1\over x}\times {1\over x} \right ) + log \ x\times {-1\over x^2}{/tex}

{tex} => {dv\over dx} = x^{1\over x} {1\over x^2}\left [1 - log \ x \right ]{/tex}   ........... (3)

From (2) and (3) we get 

{tex}=> {dy\over dx}= x^x(1 + log \ x) + x^{1\over x} {1\over x^2}\left [1 - log \ x \right ]{/tex}