How to find first derivative of function y=xln⁡(x) by limit definition, that is using this formulay′=limh→0f(x+h)−f(x)h

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Charles Benedict · Accepted Answer

limh→∞(x+h)ln⁡(x+h)−xln⁡xh=limh→∞xln⁡(x+hx)+hln⁡(x+h)h  =xln⁡[limh→∞(1+xh)1h]+limh→∞ln⁡(x+h)=1+ln⁡x  where xln⁡[limh→∞(1+xh)1h]=1 follows from the well known limit:  limh→∞(1+xh)1h=e1x

Karen Robbins · Accepted Answer

We have that  limh→0(x+h)log⁡(x+h)−xlog⁡xh=limh→0x(log⁡(x+h)−log⁡x)+hlog⁡(x+h)h=   =limh→0(x⋅log⁡(x+h)−log⁡xh+log⁡(x+h))=x⋅1x+log⁡x  indeed  log⁡(x+h)−log⁡xh=1xlog⁡(1+hx)hx→1x  indeed by y=xh→∞    log⁡(1+hx)hx=log⁡(1+1y)y→log⁡e=1

Vasquez · Accepted Answer

Continuing with the suggestion in my comment, you might do limh→0(x+h)ln⁡(x+h)−xln⁡xh=limh→0(x+h)ln⁡(x+h)−xln⁡(x+h)+xln⁡(x+h)−xln⁡xh=(limh→0ln⁡(x+h))(limh→0(x+h)−xh)+(limh→0x)(limh→0ln⁡(x+h)−ln⁡xh)=ln⁡xlimh→0(x+h)−xh+xlimh→0ln⁡(x+h)−ln⁡xh x+h−x=h , so the first limit is 1.

How to find first derivative of function y=x \ln(x) by

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