Everybody knows that derivative of x^x is x^x(1+ logx)

you would have wasted some time to differentiate x^x by taking log on both sides and stuff like that .............
but there is an easy method by which derivatives of functions of the form x^x can be found
.............
we know that
derivative of x^n is n*x^(n-1)

derivative of a^x is a^x*loga

so to differentiate x^x directly
 first differentiate x^x treating the exponent as a constant +differentiate x^x treating base as a constant .........

d(x^x) =  x*x^(x-1) [treating exponent as a constant]   +   x^x*logx[treating base as a constant]

d(x^x) = x^x + x^x*logx = x^x(1+logx) 

which is the required result........

another example

d(sinx^cosx) = cosx*sinx^(cosx-1)*cosx [treating exponent as a constant and using chain rule]  +  sinx^cosx*logsinx*(-sinx) [ treating base as a constant and using chain rule]

d(sinx^cosx) = (sinx^cosx)*{ cotxcosx  -  sinx*logsinx }


thus using this algorithm we can differentiate any exponential function with much ease without taking log and all

hope this method saves time in the exams

NOTE: THIS METHOD CAN BE USED TO CHECK THE ANSWER IN OUR CBSE BOARD EXAMS.........

THIS METHOD WAS TAUGHT BY MY IIT MATHS PROFESSOR MR.T.R.SUBRAMANIAM.........

please rate if you find it useful.................