How do you find the slope of the line tangent to the graph of #y = x ln x# at the point ( 1,0 )?

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Answer 1 · 2016-11-14T14:07:56

# y = x-1 #

The slope of the tangent at any particular point is given by the derivative.

We have #y=xlnx#

Differentiating wrt #x# using the product rule gives us:

# dy/dx=(x)(d/dxlnx) + (d/dxx)(lnx) #
# :. dy/dx=(x)(1/x) + (1)(lnx) #
# :. dy/dx=1 + lnx #

So, At# (1.0), dy/dx=1+ln1=1 #

So the required tangent passes through #(1.0)# and has slope #1#
Using #y-y_1=m(x-x_1)# the required equation is;
# y - 0 = 1(x-1) #
# :. y = x-1 #

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