How do you find the equation of tangent line to the curve #y = x^2e^-x#, with the point (1, 1/e)?

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Answer 1 · 2017-10-13T07:38:25

#y=x/e#

#y=x^2e^(-x)# can be written in the form of #y=uv# where #u# and #v# are both functions of #x#.

#(dy)/(dx)=vu'+uv'#

#v'=-e^(-x)#

#u'=2x#

#(dy)/(dx)=2xe^(-x)-x^2e^(-x)#

Putting in our #x# value of 1 gives us:

#(dy)/(dx)=2e^(-1)-e^(-1)=e^(-1)#

The gradient is #1/e#

To find the #y#-intercept we put our gradient, an #x#-value and corresponding #y#-value into #y=mx+c#

#1/e=1*1/e+c#

#c=1/e-1/e=0#

#y=x/e#