can someone help me check my work for this problem: y^2(y^2 -4) = x^2(x^2 -5) ...implicitly differentiate and then find equation of tangent line at point (0, -2)

Question

anonymous · Answer

when i worked it out i got y'=(8x^3 -10x)/(8y^3 - 8y)

anonymous · Answer

then i plugged in x for all x values and y for all y values to get the value for slope to use the point-slope formula....i ended up getting y = 2 because the numerator was = 0

turingtest · Answer

$$y^4-4y^2=x^4-5x^2$$ take the derivative w/respect to x:
$$4y^3y'-8yy'=2x^3-10x=y'(4y^3-8y)$$
$$y'=(2x^3-10x)/(4y^3-8y)$$

turingtest · Answer

plug in the values

anonymous · Answer

hmm...i wonder how i got such different values...when i first started i distributed the y^2 and the x^2 to the terms in the parenthesis..would that make a difference?

anonymous · Answer

i ended up taking derivative of: 2y^4 - 4y^2 for the y^2(y^2 -4) part

anonymous · Answer

after i distributed that y^2 to those terms, i took der. by the sum rule

turingtest · Answer

That was my first step too, but I don't see how you got 2y^4; where did the 2 coefficient come from?
 As you can see though at (0,-2) we still get y'=0
y+2=0(x-0)
y=-2
so your answer seems right to me anyway.

anonymous · Answer

ahhhh i distributed it incorrectly haha! thanks for the help im glad i asked to have someone review it haha, thanks again!

turingtest · Answer

oh yeah, and you had y=2 when it should be y=-2
Anytime!

anonymous · Answer

thanks =D