How do you find the equation of the tangent line to the curve $y = (x^{2}) e^{x + 2}$ $y=(x^2)e^(x+2)$ at x=2?

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Answer 1 · 2015-06-18T12:32:36

Find the slope of the tangent line using the derivative, find the $y$ $y$ value of the point where $x = 2$ $x=2$ , and find the equation of the line.

$y = (x^{2}) e^{x + 2}$ $y=(x^2)e^(x+2)$

at x=2, we get $y = 4 e^{4}$ $y=4e^4$ , So our line goes through the point: $(2, 4 e^{4})$ $(2, 4e^4)$

To find the slope, find $y'$ . To find $y'$ use the product rule and the derivative of exponentials (and the chain rule).

$y' = 2xe^(x+2) + x^2 e^(x+2) (1)$ -- I included the $(1)$ to remind us of the chain rule.)

At $x=2$ , $m = y' |_2 = 4e^4+4e^3 = 8e^4$

The line with $m=8e^4$ through $(2, 4e^4)$ has equation(s):

$y-4e^4 = 8e^4(x-2)$

$y-4e^4 = 8e^4x-16e^4$

$y = 8e^4x - 12 e^4$

How do you find the equation of the tangent line to the curve y=(x^2)e^(x+2)y=(x2)ex+2y=(x^2)e^(x+2) at x=2?