Question

Use implicit differentiation to find equation of tangent line to the curve at the given point.

Answer 1

\[y^2(y^2-4)=x^2(x^2-5)\] point: ( 0, 2) (Devil's curve)

Answer 2

\[[(y^2-4) \times 2y \times y' + y^2 \times 2y \times y' ] = (x^2-5) \times 2x + x^2 \times (2x)\]

Answer 3

\[2y(y^2-4)y' + 2y^3y' = 2x(2x^2-5)\]

Answer 4

thats what I have so far, I don't know if its correct.

Answer 5

Thank you for your time. I figured it out. :)

Answer 6

\[y^2(y^2-4)=x^2(x^2-5)\]\[y^4-4y^2=x^4-5x^2\]\[\frac{d}{dx}[y^4-4y^2=x^4-5x^2] \implies 4y^3 \cdot y' -8y \cdot y' =4x^3-10x\]\[y'(4y^3-8y) = 4x^3-10x\]\[y'=\frac{4x^3-10x}{4y^3-8y} = \frac{2(2x^3-5x)}{4(y^2-2y)} = \frac{2x^3-5x}{2(y^2-2y)} \] That's what I'm getting.

Answer 7

Alright