Question

Find the equation of the tangent line to the curve (a lemniscate) 2(x^2+y^2)2=25(x^2−y^2) at the point (−3,1). The equation of this tangent line can be written in the form y=mx+b where m is:

Answer 1

There's an implicit derivative in your future.

Answer 2

yes, I know that part. do i need to simp;ify first then F'?

Answer 3

Why? If you solve first, it won't be implicit.

2*2(x^2 + y^2)(2x + 2yy') = 2x - 2yy'

Now solve for y'.

Answer 4

how did you get this?

Answer 5

where did 25 go?

Answer 6

I lost it. Please put it back. Sorry about that. End of a line confusion.

Answer 7

\[f(x)=2(x^{2}+y^{2})^2=25(x^{2}-y^{2})\]

Answer 8

This is the equation :9

Answer 9

No. it's not. You invented the f(x). The implicit derivative is exactly as I stated it, excepting the return of the 25.

2*2(x^2 + y^2)(2x + 2yy') = 25(2x - 2yy')

Solve for y'.

Answer 10

why its 2*2 in the front?

Answer 11

Im confused. how did you get your equation?

Answer 12

It's an implicit dervative. Assume y = f(x) and use lots of chain rule.

Answer 13

Have you done an implicit derivative?

Answer 14

prudct rule?

Answer 15

No, i am not sure.

Answer 16

(d/dx)(xy) = xy' + y -- Product and Chain Rule

(d/dx)(y^2) = 2yy' = Power and Chain Rule

Keep in mind that we are assuming y = f(x).

Answer 17

Yes! thanks for the fact.
this is what exactly I do not understand.
product and chain rule. how do i find out?
I know the rule but when it comes to double or mixed of two rules I get confused. I have test next week, please help me

Answer 18

Well, if this is your first exposure to the implicit derivative, we are in trouble. I have no confidence that we can cover the required territory in this forum.