Curve has equation y = `e^(2x)/(x+3)^2`   Show that =  Ae^2x(x+2)/(x+3)^3 Where A is a constant to be found  And find the exact coordinate of point on the curve where `dy/dx=0`  

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gsenviro eNotes educator| Certified Educator

equation of the given curve: `y= e^(2x)/(x+3)^2`

The derivative of this equation, dy/dx can be calculated by using quotient rule.

i.e., `d/dx f(x)/g(x) = (g(x)f'(x) - f(x)g'(x))/(g(x))^2`

Here, f(x) = `e^(2x)` and f'(x) = `2e^(2x)`

and g(x) = `(x+3)^2`  and g'(x) = 2(x+3) 

Therefore, `dy/dx = ((x+3)^2 2e^(2x) - e^(2x) 2(x+3))/((x+3)^2)^2`  

using the quotient rule and substituting the values of f(x), f'(x), g(x)and g'(x)

simplifying the equation, we get

`dy/dx = (C(x+3) 2e^(2x) (x+3-1))/(x+3)^4 = (Ae^(2x) (x+2))/(x+3)^3`

where, C and A are constants and their values can be found given other conditions.

For dy/dx = 0, x= -2 , since e^2x can not be 0 and A is non-negative as well, so (x+2) = 0, which means, x = -2

and correspondingly, y = `e^(2*-2) / (-2+3)^2 = e^(-4)`

Therefore, dy/dx = 0 at (-2, e^-4).

Hope this helps.