Question

Finding an Equation of a tangent Line in Exercise, find an equation of the tangent line to the graph of the function at the given point.
y = x/e^2x , (1, 1/e^2)

Answer 1

Answer:

$y=-\dfrac{x}{e^{2}}+\dfrac{2}{e^{2}}$

this is the equation of the tangent line to the curve from the point (1, 1/e^2)

Step-by-step explanation:

$y=\dfrac{x}{e^{2x}}$

to find the tangent line we need to find the curve's derivative.

we'll be using the quotient formula: $d\left(\dfrac{u}{v}\right) = \dfrac{uv'-vu'}{v^2}$, here $u = x\quad , \quad v = e^{2x}$

$\dfrac{dy}{dx}=\dfrac{e^{2x}(1)-x(2e^{2x})}{(e^{2x})^2}$

$\dfrac{dy}{dx}=\dfrac{e^{2x}-2xe^{2x}}{(e^{4x})}$

$\dfrac{dy}{dx}=\dfrac{\left(1-2x\right)}{ e^{2 x}}$

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

$\dfrac{dy}{dx}=\dfrac{\left(1-2(1)\right)}{ e^{2(1)}}\\\dfrac{dy}{dx}=\dfrac{-1}{e^{2}}$

this is the slope of the tangent.

now to find the equation of the line:

$(y-y_1)=m(x-x_1)$

here, m is the slope.

$(y-\dfrac{1}{e^{2}})=-\dfrac{1}{e^{2}}(x-1)\\y=-\dfrac{x}{e^{2}}+\dfrac{1}{e^{2}}+\dfrac{1}{e^{2}}$

$y=-\dfrac{x}{e^{2}}+\dfrac{2}{e^{2}}$

this is the equation of the tangent line to the curve from the point (1, 1/e^2)