Derivative of 1+logx ????

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Answer 1 · 2018-02-22T03:30:30

#1/(xln10)#

We are asked to find: #d/dx(1+log(x))#

We can use the sum rule here, which states that

#d/dx(x+y)=d/dx(x)+d/dx(y)#

We know that the derivative of a constant is #0#, therefore #d/dx(1)=0#.

So, we get

#d/dx(1+log(x))=d/dx(1)+d/dx(log(x))#

#=0+d/dx(log(x))#

#=d/dx(log(x))#

The logarithm derivative is a common one, where

#d/dx(log(x))=d/dx(log_10(x))=1/(xln10)#

#:.d/dx(1+log(x))=1/(xln10)#