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Defferenentiate of x^x = ...?

____________________________________

Answer
     x
    x  (1 + lnx)
Explanation

1) Here given question not a power functions of form x^a so we can't apply the power formula of integration.

     d    n      n-1
    ---- x  = n.x
     dx  

2) Here also given question is also not an exponent function of form a^x so we can't apply the exponent formula of integration.

     d    x    x
    ---- a  = a  . ln a
     dx  

Now 
     d    x    
    ---- x  
     dx  
          
         x
Let y = x

Take natural logarithms both sides. 
               x
    ln y = ln x

    ln y = x.ln x

Differenetiate with respect to x. 

            d            
    ln y = ---- (x.ln x) 
            dx           

     1   dy       d                 d
    --- ---- = x.---- (ln x) + lnx.----(x)
     y   dx       dx                dx

     1   dy       1                
    --- ---- = x.--- + lnx.(1)
     y   dx       x                

     1   dy                     
    --- ---- = 1 + lnx
     y   dx                     

     dy                     
    ---- = y(1 + lnx)        Put value of y
     dx              

     d    x   x                
    ---- x = x  (1 + lnx)
     dx              

____________________________________

Note

1) Before blidly using any formula of integraion becareful about x^x is not power formula like x^a or exponential function like a^x. So apply we can't apply the power formula or exponential formula of integration.

2) Here x^x both base and exponent are variables. Remember it always in such case we should solve it by applying logarithm both sides.

3) Natural logarithms represt by ln so dont use 'log' sign

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