Find the equation of the tangent line to the curve `y=xe^(2x)` at the point `(2,2e^4)`

Expert Answers
lemjay eNotes educator| Certified Educator

Take note that the derivative of the curve is the slope of the tangent line at the point of tangency.

So, let's take the derivative of the curve.



Use the product rule of derivatives to determine y'. The formula is:


Then, let

`u=x `                   and               `v=e^(2x)`

`u'=1`                                      `v'= e^(2x)*2=2e^(2x)`

So, we have

`y'=x*2e^(2x)+e^(2x)*1 = 2xe^(2x)+e^(2x)=e^(2x)(2x+1)`

Note that y' contains the x variable only. So, substiute the x-coordinate of the point of tangency `(2,2e^4)` .


Hence, the slope of the tangent line is:


Then, substitute the point `(2,2e^4)` and `m=5e^4` to the point-slope form of a line.

`y-y_1 = m (x-x_1)`





Hence, the equation of tangent line is `y=5e^4x-8e^4` .

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