Question

Given y-2xy=x^3, determine the equation of the tangent line where the point of tangency occurs at x=3

Answer 1

I'll be right back tomorrow. I have to go to sleep. If someone can help me start the question that will be right. Appreciated it! :)

Answer 2

You need a `point` on the line and `slope` to write the equation of tangent line

Answer 3

derivative at x=3 gives the slope of tangent line at x=3

Answer 4

y-2xy = x^3

y(1-2x) = x^3

y = x^3/(1-2x)

dy/dx = ?

Answer 5

@ganeshie8
i need help to figure out dy/dx

Answer 6

You have : \[y = \dfrac{x^3}{1-2x}\]
maybe use quotient rule to find \(\dfrac{dy}{dx}\)

Answer 7

okay, let me try it

Answer 8

\[y'=\frac{ -4x^3+3x^2 }{ (1-2x)^2 }\]
is that correct?

Answer 9

Yes! plugin x=3 since you want the slope of tangent line at x=3

Answer 10

is it -27/5?

Answer 11

@ganeshie8 how do i figure the equation?

Answer 12

doesn't look correct, check again

Answer 13

wolfram says -81/25
http://www.wolframalpha.com/input/?i=%28x%5E3%2F%281-2x%29%29%27+at+x%3D3

Answer 14

oh i got -81/25 now @ganeshie8

Answer 15

good so you know the "slope" of line
you just need a point on the line to write the equation, yes ?

Answer 16

yes!

Answer 17

plugin x=3 in to the given equation and solve y

Answer 18

y-2xy=x^3

plugin x=3

Answer 19

y= -27/5?

Answer 20

Yep so you have :

slope of line = -81/25
point on line = (3, -27/5)

Answer 21

oh so it will -81/25=(y+27/5)/(x-3)

Answer 22

Looks good!

Answer 23

so it would be 0=81x-25y-108?

Answer 24

thanks! @ganeshie8