Question

A bit confused in this implicit differentiation problem: 3(x^2+y^2)^2=100(x^2-y^2)
6(x^2+y^2)(2x+2yy')=100(2x-2yy')
Would I just foil 6(x^2+y^2)&(2x+2yy') out, then factor out the y'? And then move everything without y' to the right?

Answer 1

I believe that is correct.

Answer 2

I thought so too, but I checked the problem and that is shown.

So, in all the other problems I found y' and did it as normal to get to the equation of the tangent line(find y', input points (x,y), point slope form), but apparently not this one. Is there any particular reason as to why they just added the points in before actually finding y'?

Answer 3

Because once you derive the equation, it is not important that you continue using the variable x, and y, because once you have the function of the derivative, and all you need is to solve for y'. You need to differentiate the equation before inserting the points though because you need the y' in there. But after you derive it, you can just sub x, y, points and solve for y'. This makes it a little less bothersome because you can simplify it with numbers instead of carrying x and y around the whole time when you are trying to isolate y'.

Answer 4

Just remember, you can only sub in the points After you differentiate it

Answer 5

I totally agree with you! But, why didn't they do it there?
6(x^2+y^2)(2x+2yy')=100(2x-2yy')
Is it?

Answer 6

Wait, so I can sub the points in only after I differentiate all the terms? Then solve for y'?
and that would still give me my tangent slope?

Answer 7

Yes

Answer 8

Ex: x^2y^2-9x^2-4y^2=0 original equation
Points (-4,2 sqrt3)

x^2yy'+2y^2x-18x-8yy'=0 at this point I could sub the points in and solve for y'?

Answer 9

I get 2x*y^2+x^2*2y*y'-18x-8y*y'=0 as the derivative function, but anyway, yes, after you take the derivative, you can just sub in the x, y, points and solve for y' and it would be the same as if you didn't sub it in early

Answer 10

Dude, you just saved me a lot of time. Ily. thanks!