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

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### 1 Answer

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.

`y=xe^(2x)`

`y'=d/dx(xe^(2x))`

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

`d/dx(uv)=uv'+vu'`

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)` .

`y'=e^(2*2)(2*2+1)=e^4(5)=5e^4`

Hence, the slope of the tangent line is:

`m=y'=5e^4`

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)`

`y-2e^4=5e^4(x-2)`

`y-2e^4=5e^4x-10e^4`

`y=5e^4x-10e^4+2e^4`

`y=5e^4x-8e^4`

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

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