Question

Write an equation for the tangent line to the graph @ (1,1) 4xy = (x^2 +y^3)^2

Answer 1

\[\large 4xy=(x^2+y^3)^2\]

Hmm so we need to start by finding the derivative.
Have you tried doing that yet? :)

Answer 2

yea.. i end up with

4(x'y+xy') = 2(x^2 +y^3) + (2x +3y^2 y')

Answer 3

insert the x,y values...
i get 4(1+y') = 2(1 +1) + (2[1]+3[1}y')

Answer 4

Hmm ok I see one small little thing we should fix,\[\large 4(1+y')=2(1+1)(2+3y')\]Chain rule tells us to `multiply`. So we gotta get rid of that extra addition sign there.
Ok looks good so far.
So now solve for y'. Understand how to do that?

Answer 5

thats wat i thought.. in class professor added the addition sign and it through me off.

Answer 6

Thank you mate.

Answer 7

Oh got it from there? c: ok cool.