Question

I really need help!
Find the equation of the line tangent to the curve (x^2 + y^2)^2 = 4xy

Answer 1

just gotta take the derivative of that

Answer 2

say that each variable is a function of a common unknown variable; say, t

(x(t)^2 + y(t)^2)^2 = 4x(t)y(t)

now you should already know the power rule, the chain rule and the product rules

Answer 3

I still don't understand.. :S

Answer 4

do you know how to take a derivative?

Answer 5

Yes, I know the power rule! But I'm not sure how to do it for this particular question. Would it be:
4x^3+4y^3=4xy ??

Answer 6

\[D[ (x^2 + y^2)^2 = 4xy]\]
\[D[ (x^2 + y^2)^2 = D[4xy]\]
notice the power rule, and the product rule, and the chain rule that this produces
\[ 2(x^2 + y^2)~D[x^2 + y^2] = 4(D[x]y]+xD[y])\]
\[ 2(x^2 + y^2)~(D[x^2] + D[y^2]) = 4(x'y+xy')\]
\[ 2(x^2 + y^2)~(2xx' + 2yy') = 4(x'y+xy')\]
and simplify
\[(x^2 + y^2)~(xx' + yy') = x'y+xy'\]

dy/dx = y' ; and dx/dx = x' = 1

mnost of this work is just in figuring out the algebra to move things around with.

\[(x^2 + y^2)~(x + yy') = y+xy'\]
\[x^2x + x^2yy'+xy^2 + y^2yy' = y+xy'\]
\[x^2yy' + y^3y'-xy' = y-x^3-xy^2\]
\[y'(x^2y + y^3-x) = y-x^3-xy^2\]
\[y' = \frac{y-x^3-xy^2}{x^2y + y^3-x}\]