Respuesta :
Answer:
[tex]y=-\dfrac{x}{e^{2}}+\dfrac{2}{e^{2}}\\[/tex]
this is the equation of the tangent line to the curve from the point (1, 1/e^2)
Step-by-step explanation:
[tex]y=\dfrac{x}{e^{2x}}[/tex]
to find the tangent line we need to find the curve's derivative.
we'll be using the quotient formula: [tex]d\left(\dfrac{u}{v}\right) = \dfrac{uv'-vu'}{v^2}[/tex], here [tex]u = x\quad , \quad v = e^{2x}[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{e^{2x}(1)-x(2e^{2x})}{(e^{2x})^2}[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{e^{2x}-2xe^{2x}}{(e^{4x})}[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{\left(1-2x\right)}{ e^{2 x}}[/tex]
This is the equation of the slope of the curve. By finding the slope of the curve at (1, 1/e^2) we'll also be finding the slope of the tangent to the curve at (1, 1/e^2).
the x-coordinate is 1, so using x =1
[tex]\dfrac{dy}{dx}=\dfrac{\left(1-2(1)\right)}{ e^{2(1)}}\\\dfrac{dy}{dx}=\dfrac{-1}{e^{2}}[/tex]
this is the slope of the tangent.
now to find the equation of the line:
[tex](y-y_1)=m(x-x_1)[/tex]
here, m is the slope.
[tex](y-\dfrac{1}{e^{2}})=-\dfrac{1}{e^{2}}(x-1)\\y=-\dfrac{x}{e^{2}}+\dfrac{1}{e^{2}}+\dfrac{1}{e^{2}}\\[/tex]
[tex]y=-\dfrac{x}{e^{2}}+\dfrac{2}{e^{2}}\\[/tex]
this is the equation of the tangent line to the curve from the point (1, 1/e^2)
