Find the slope of the tangent line to the curve (a lemniscate)

Question

2(x^2+y^2)^2=25(x^2–y^2) at the point
(–3,1) .

I know that we are supposed to use chain rule for this problem but i think i am messing up on the algebra. Can someone please help me, i keep on getting m= (42x^3-42y^2x)/(58x^2+58y^2)

Answer 1

2(x^2+y^2)^2=25(x^2–y^2)
2(x^2+y^2)^2-25(x^2–y^2)=0
Expand
2x^4+4x^2y^2+2y^4-25x^2+25y^2=0
Differentiate (implicitly) with respect to x, denote dy/dx as y'
8x^3+8xy^2+4x^2(2y)y'+8y^3y'-50x+50yy'=0
Collect terms and solve for y' in terms of x and y:
y'=(50x-8x^3-8xy^2)/(8x^2y+8y^3+50y)
At (-3,1), x=-3, y=1
y'=(-150+216+24)/(72+8+50)=9/13

Check for arithmetic errors.

Answer 2

I get
4(x^2 + y^2)(2x + 2yy') = 50x - 50y y'
.....
....
y' = ((50x - 8x^3 - 8xy^2)/[ (8x^2)y + 8y^3 - 50y)

plug in your point (-3,1) to get the slope

Take it from there.

Answer 3

Oh my goodness, thank you soooo much for writing it out i totally wasn't getting it! THank you so much :)

Do you think you coul check my arithmatic for another problem? It's
Use implicit differentiation to find the slope of the tangent line to the curve

y/x+6y=x^2–6 at the point (1–5/31)

I got [(x+6y)(y')-(y)(1+6y')]/(x+6y)^2=2x then y'=(2x+x^2+12yx+6y^2-y)/(x+6y+6). Is that right?

Answer 4

y/(x+6y) or (y/x)+6y?
As it is, the expression means the latter.

Since I see (x+6y)y' somewhere, I think you mean the first interpretation.

y/(x+6y)=x^2–6
Cross multiply to get
y=(x^2-6)(x+6y)
y'=2x(x+6y)+(x^2-6)(1+6y')
I get
y'=-(12xy+3x^2-6)/(6x^2-37)
Check my arithmetic and differentiation and take it from here.