What is the derivative of # y= xlnx#?

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Answer 1 · 2017-05-16T17:33:21

# dy/dx = 1 + lnx #

We have:

# y= xlnx#

We can apply the product rule to get:

# dy/dx = (x)(d/dx lnx) + (d/dxx)(lnx) #

Noting a standard calculus result:

# d/dx lnx = 1/x #

We get:

# dy/dx = x*1/x + 1*lnx #

# " " = 1 + lnx #

Corollary
We have just shown that:

# d/dx (xlnx) = 1 + ln x #

If we now integrate both sides, then we get:

# \ \ \ \ \ xlnx = int \ (1 + ln x) \ dx#

# :. xlnx = x -c + int \ ln x \ dx #

Hence:

# int \ ln x \ dx = xlnx -x +c#