Posted by jasmineT on Saturday, February 18, 2012 at 10:00pm.
Find the slope 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) .
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^342y^2x)/(58x^2+58y^2)

Implicit Differentiation  MathMate, Saturday, February 18, 2012 at 10:15pm
2(x^2+y^2)^2=25(x^2–y^2)
2(x^2+y^2)^225(x^2–y^2)=0
Expand
2x^4+4x^2y^2+2y^425x^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'=(50x8x^38xy^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.

Implicit Differentiation  Reiny, Saturday, February 18, 2012 at 10:21pm
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.

Implicit Differentiation  jasmineT, Saturday, February 18, 2012 at 10:30pm
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^2y)/(x+6y+6). Is that right?

Implicit Differentiation  MathMate, Sunday, February 19, 2012 at 8:17am
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^26)(x+6y)
y'=2x(x+6y)+(x^26)(1+6y')
I get
y'=(12xy+3x^26)/(6x^237)
Check my arithmetic and differentiation and take it from here.
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