What is the derivative of # y= xlnx#?
1 Answer
May 16, 2017
# dy/dx = 1 + lnx #
Explanation:
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#