Question

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

Answer 1

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

Explanation:

`•color(white)(x)dy/dx=m_(color(red)"tangent")" at x = a"`

`"differentiate using the "color(blue)"product rule"`

`"given "y=g(x).h(x)" then "`

`dy/dx=g(x)h'(x)+h(x)g'(x)larr" product rule"`

`g(x)=x^2rArrg'(x)=2x`

`h(x)=e^((x+2))rArrh'(x)=e^((x+2))`

`rArrdy/dx=x^2e^((x+2))+2xe^((x+2))`

`"at x = 2"`

`dy/dx=4e^4+4e^4=8e^4`

`" and " y=4e^4`

`"using "m=8e^4" and "(x_1,y_1)=(2,4e^4)`

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

`rArry=8e^4x-12e^4`