Question

Determine the equation of the tangent line at the given point.

Answer 1

\[y=\frac{ x }{ e ^{2x} }\] at \[(1, \frac{ 1 }{ e ^{2} }\]

Answer 2

ok... so whats the 1st derivative..?

Answer 3

the 1st derivative is the equation for the slope of the tangent at any point on the curve.

so when you have the 1st derivative substitute x = 1 to find the slope at your nominated point.

then you have the point and slope so you can find the equation of the tangent.

Answer 4

but then what about the value of y coordinate? does it not matter?

Answer 5

well before you worry about the y value work on the 1st derivative....

you will use the y value in the point to find the equation of the line...

\[y - y_{1} = m(x - x_{1})\]

if you use the formula above x = 1 and y = e^-2

but before that its

1. 1st derivative

2. the value of the slope after substitution.

Answer 6

ok so for the first derivative i got: \[\frac{ 1-x }{ e ^{2x} }\]

Answer 7

Is that right?

Answer 8

I got

\[\frac{ 1 - 2x}{e^{2x}}\]

Answer 9

ah forgot the 2 while doing chain rule. yeah you're right.

Answer 10

ok... so now substitute x =1 into the derivative and you will have the slope of the tangent..

I'd leave the answer as an exact value.

does that make sense...

Answer 11

yes but when i substitute 1 i get -1/e^2, when the answer is positive, not negative...

Answer 12

yep thats great now well use the x and y values for the point as well as the slope..

\[y - \frac{1}{e^2}= \frac{-1}{e^2}(x - 1)\]

just simplify to find the equation of the tangent.

Answer 13

ok got it. thanks!

Answer 14

so this is what you are looking at

hope it helps

Answer 15

thanks a ton! :D