Question

differentiate y^3+xy+2=0
(find dy/dx)

answer is -2/7 (i think) but not sure how to get there, i got -1/3

can someone differentiate and show the work quick?

Answer 1

Is this the chapter on implicit differentiation?

Answer 2

yeah

Answer 3

kk

Answer 4

\[3y^2y'+ xy'+y=0 \]

Answer 5

brb sec, i have a question being answered.

Answer 6

i got f'(x)= 3y^2(dy/dx)+y+x(dy/dx)
0 =

-y= dy/dx (3y^2+x)

then divided by (3y^2+x)

so got dy/dx = -y/3y^2+x is that right?

Answer 7

thats what I just got.

Answer 8

but then the next question is to find the tan line approx to the curve at the point (-5,2)

Answer 9

plug in the values

Answer 10

and i can't do that if i dont have an m value (which i think is found in the last question)

Answer 11

m is your y'

Answer 12

(y-2) = y'(x+5)

Answer 13

ohh got it. it wasn't clicking tht the y' was dy/dx

Answer 14

thanks