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Implicit Differentiation

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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^3-42y^2x)/(58x^2+58y^2)

  • Implicit Differentiation -

    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.

  • Implicit Differentiation -

    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 -

    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?

  • Implicit Differentiation -

    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.

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