What is the derivative of #xlogx#?

2 Answers
Truong-Son N.
Jun 15, 2017

#d/(dx)[xlogx] = log(ex)#.

Assuming you mean #xlog_(10)x#... the derivative of #lnx# is #1/x#, so using the change of base law:

#d/(dx)[log x] = d/(dx)[(logx)/(log 10)] = d/(dx)[(lnx)/(ln10)]#

#= 1/(xln10)#

Therefore, using the product rule, where

#d/(dx)[f(x)g(x)] = f(x)(dg)/(dx) + g(x)(df)/(dx)#,

we obtain, using the derivative of #logx# we got earlier:

#color(blue)(d/(dx)[xlogx]) = x(d(logx))/(dx) + logx cancel((d(x))/(dx))^(1)#

#= cancel(x)/(cancel(x)ln10) + logx#

#= 1/(ln10) + logx#

#= (log e)/(log 10) + logx#

#= log e + log x#

#= color(blue)(log(ex))#

Shwetank Mauria
Jun 15, 2017

#d/(dx)xlogx=0.4343(1+lnx)#

Explanation:

#f(x)=xlogx=xlnx/ln10=1/ln10xlnx=0.4343xlnx#

Hence #(df)/(dx)=0.4343(x xx 1/x+lnx)#

= #0.4343(1+lnx)#

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