How would you solve for the slope of the line tangent to the curve (y^2)+(xy+1)^3 = 0 at (2,-1) using implicit. the answer should be 3/4 thanks :)

Question

johan14th · Answer

$$y^2+(xy+1)^3 = 0 $$

irishboy123 · Answer

have you tried doing it?

take each term as it comes.

johan14th · Answer

$$2y\frac{ dy }{ dx } + 3(xy+1)^2 (y+x \frac{ dy }{ dx })=0$$

johan14th · Answer

$$\frac{ dy }{ dx } = \frac{ -y(3xy^2+3) }{ 2y+x }$$

johan14th · Answer

this is what i got but when i try to evaluate it at ( 2,1) i cant get the right answer. Is there something i did wrong?

irishboy123 · Answer

the answer is right.  i think your tangent is wrong.  i've dome it a different way and get $y' = -\dfrac{3(xy+1)^2y}{2y + 3(xy+1)^2 x}$

irishboy123 · Answer

yeah, look at your algebra, your first line seems ok

irishboy123 · Answer

$2y\frac{ dy }{ dx } + 3(xy+1)^2 (y+x \frac{ dy }{ dx })=0$
$2y\frac{ dy }{ dx } + 3(xy+1)^2 y+ + 3(xy+1)^2 x \frac{ dy }{ dx }=0$

johan14th · Answer

okay i think i have discovered my mistake when i moved the equation from the left to the right

irishboy123 · Answer

yes!

johan14th · Answer

i just saw it when you wrote it. thank you!

irishboy123 · Answer

mp

:p