Question

Use implicit differentiation to find an equation of the tangent line to the curve at the given point 2(x^2+y^2)^2=25(x^2-y^2) and the point (3,1)

Answer 1

have you gotten started on this at all, or are you totally lost?

Answer 2

I started by doing the power rule and getting (4x^2+4y^2)[2x+2y dy/dx]=25(2x-2y dy/dx) and then distributed,isolated dy/dx, etc, but my answer is wrong because they are getting a negative answer and from the derivative I get I wouldn't get a negative answer.

Answer 3

I didn't mean to post it in both places, but thank you so much for answering!

Answer 4

ok

Answer 5

Could I do it with the power and chain rule as well? or no?

Answer 6

yes, you could, just carefully take derivative with y.

Answer 7

you can simply use the chain rule and isolate the dy/dx component from which you have the gradient where you can substitute x=3 to have the gradient at that point and so, using the formula (y-1)=m(x-3) where m is the gradient you have calculated which is then the equation of the tangent at (3 ,1).

Answer 8

from where i think the gradint is positive

Answer 9

an i think m=45/17