Question

The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is

Answer 1

the implicit derivative

Answer 2

is that when you take the derivative and include y' ?

Answer 3

yes, which afterall is the same as taking a regular derivative, only "y" isnt easy to pull out

Answer 4

right. That's where I went wrong. I tried to take the derivative, but I must have messed up somewhere because I got a really odd number that's not one of my answer choices.

Answer 5

once I get the derivative, can I plug in immediately for x and y, or do I have to reduce / combine?

Answer 6

\[\frac{d}{dx}(y^2 + (xy+1)^3 = 0)\]
\[\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0\]
\[\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0\]
\[\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0\]
\[\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0\]
\[\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0\]
\[\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0\]
\[\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right) (xy+1)^2 = 0\]
this gets messy lol

Answer 7

once you get the derivate, use the point to calibrate it, yes

Answer 8

I got 3/4

Answer 9

\[2y\ y' + \left(3y+3x\ y'\right) (xy+1)^2 = 0\]
\[2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0\]
\[2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0\]
\[2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0\]
\[3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0\]
\[y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y\]
\[y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1\]
\[y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}\]
\[y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}\]
\[y' = \frac{12 -12+3}{(6 -24-12-2 )}\]
\[y' = \frac{3}{( -32 )}\]
but i couldve missed it along the way