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

Question

zenmo · Answer

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

zenmo · Answer

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

zenmo · Answer

$$2y(y^2-4)y' + 2y^3y' = 2x(2x^2-5)$$

zenmo · Answer

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

zenmo · Answer

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

jhannybean · Answer

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

jhannybean · Answer

Alright