How do you find the equation of the tangent line to the graph #y=e^-xlnx# through point (1,0)?

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Answer 1 · 2017-03-19T00:52:59

# y = 1/ex-1/e #

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

We have:

# y = e^(-x)lnx #

First let us check that #(1,0)# lies on the curve:

# x=1 => y=1/eln1 = 0 #

Then differentiating wrt #x# (using the product rule) gives us:

# dy/dx = (e^(-x))(1/x) + (e^(-x))(lnx) #
# " " = e^(-x)(1/x+lnx) #

When #x = 1 => dy/dx = 1/e(1+ln1) =1/e #

So the tangent passes through #(1,0)# and has gradient #1/e# so using the point/slope form #y-y_1=m(x-x_1)# the equation we seek is;

# y-0 = 1/e(x-1) #
# :. y = 1/ex-1/e #

We can confirm this solution is correct graphically:
enter image source here