How do you find the equation of the line tangent to #y=(x^2)(e^3x)# at the point where x=1/3?

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Answer 1 · 2016-03-21T13:38:21

Equation of tangent is #27y-9e^3x+2e^3=0#

As #y=(x^2)(e^3x)=e^3x^3# hence at #x=1/3#, #y=e^3(1/3)^3=e^3/27# i.e. curve passes through #(1/3,e^3/27)#.

Now #(dy)/(dx)=3e^3x^2# and hence as it gives the slope of the tangent at #(x,y)#,

slope is #3e^3(1/3)^2# or #e^3/3#

Hence equation of line passing through #(1/3,e^3/27)# and slope #e^3/3# is

#(y-e^3/27)=e^3/3(x-1/3)# or

#27y-e^3=9e^3x-3e^3# or

#27y-9e^3x+2e^3=0#