When differentiating an exponential function: d d x a x = a x ln ⁡ ( a ) , ( a &gt; 0 )Why does differentiating x x result in x x ( 1 + ln ⁡ x ) and not x x ln ⁡ x?

Question

When differentiating an exponential function:      d          d      x            a    x    =      a    x    ln  ⁡  (  a  )  ,    (  a  &amp;gt;  0  )Why does differentiating       x    x   result in       x    x    (  1  +  ln  ⁡  x  ) and not       x    x    ln  ⁡  x?

smeachtaczm · Accepted Answer

By definition,       x    x    =  exp  ⁡  (  x  ln  ⁡  (  x  )  ). Thus,  (      x    x        )    ′    =  (  x  ln  ⁡  (  x  )      )    ′    exp  ⁡  (  x  ln  ⁡  (  x  )  )  =  (      x    ′    ln  ⁡  (  x  )  +  x      ln    ′    ⁡  (  x  )  )  ⋅      x    x    =  (  ln  ⁡  (  x  )  +  1  )  ⋅      x    x    .

Dylan Benitez · Accepted Answer

The rule you stated only works when the base is a fixed constant like       2    x   or       7    x  . In the expression       x    x  , the base is a variable. So your quoted rule does not apply here. To differentiate the function   f  (  x  )  =      x    x   i think you understand how to do so. So I will not include that in my answer.

When differentiating an exponential function: d/dx a^x=a^xln(a),(a>0) Why does differentiating x^x result in x^x(1+lnx) and not x^x lnx?

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