Question

Consider the curve given by y^2=4xy+1. Find dy/dx. I found (-2y)/(y-2x) = dy/dx. First, if you could validate that answer, then help me with this: Find all points on the curve where the slope of the tangent line is 2.

Answer 1

I've started by setting \[-2y \div (y-2x)=2\], but I don't know where to go from there.

Answer 2

use implicit differentiation:
2y(dy/dx)=4(x*(dy/dx)+y*1) + 0
2y(dy/dx) = 4x(dy/dx)+4y

Answer 3

(2y)/(y-2x) = dy/dx.
this is the answer

Answer 4

2y(dy/dx)-4x(dy/dx)=4y
Pull out the dy/dx:
(dy/dx)(2y-4x)=4y
dy/dx = 4y/(2y-4x)
and through dividing everything by 2, you get waqassaddique's answer which is:
dy/dx = 2y/(y-2x)

Answer 5

Good, that's what I got. Can you help me with the second part? What are you supposed to do to get such an answer?

Answer 6

Find all the points where M=2 means dy/dx=2

Answer 7

Mhm.

Answer 8

so:
2 = 2y/(y-2x)
1 = y/(y-2x)
y-2x = y
-2x = 0
x=0
Point:
(0;f(0))

Answer 9

f(0) = +/-1

Answer 10

The y's just cancel out?

Answer 11

\[y^2=4xy+1\]
\[2yy'=4y+4xy'\] then algebra
\[2yy'-4xy'=4y\]
\[(2y-4x)y'=4y\]
\[y'=\frac{4y}{2y-4x}=\frac{2y}{y-2x}\] is what i get

Answer 12

yes
and this is problem

Answer 13

yea, just subtract y from both sides.
Points are:
(0;-1) and (0;1)

Answer 14

yeah bahrom has the same

Answer 15

Thank you all! :) I really appreciate it.

Answer 16

yay, high five satellite!

Answer 17

and u're welcome m.

Answer 18

set = 2 and solve yes?

Answer 19

Yes. bahrom just helped me with that. I needed help with the solving, but apparently the ys just cancel.