Question

How do you implicitly differentiate: x^3y^3-y=x

Answer 1

take y to the other side then apply product rule in the left hand side

Answer 2

gimme a sec i am working on it

Answer 3

you aint gots to move anything, just differentiate

Answer 4

haha amistre is correct

Answer 5

:) it happens

Answer 6

x^3y^3-y=x

x'^3y^3+x^3y^'3-y'=x'

Answer 7

eventually you factor out a y' and move it all to the other side tho

Answer 8

d/dx(x^3y^3)-d/dx(y)=d/dx(x)

Answer 9

For d/dx(x^3y^3) you use the Chain Rule?

Answer 10

product rule, and some say chain rule for the y(x) part

Answer 11

What are the steps please?

Answer 12

\[x^3y^3-y=x\] is like
\[x^3f(x)^3-f(x)=x\]

Answer 13

for first term you use the product rule and the chain rule
\[3x^2f^3(x)+x^33f^2(x)f'(x)\] or
\[3x^2y^3+3x^3y^2y'\]

Answer 14

so start with
\[3x^2y^3+3x^3y^2y-y'=1\] then solve for
\[y'\]

Answer 15

rather
\[3x^2y^3+3x^3y^2y'-y'=1\]

Answer 16

So the answer is: \[y' = (1-3x^2y^3)/(3x^2-1)\]

Answer 17

forgot a y^2 in the denom

Answer 18

but other than that, yes

Answer 19

So it's \[y′=(1−3x^2y^3)/(3x^2−y^2)\]

Answer 20

\[y′=\frac{1−3x^2y^3}{3x^2y^2−1}\]

Answer 21

I'm trying to figure out where I forgot the y^2..

Answer 22

\[3x^2y^3+3x^3y^2y'-y'=1\]
\[3x^3y^2y'-y'=1-3x^2y^3\]
\[y'(3x^3y^2-1) =1-3x^2y^3\]

Answer 23

Yes, the correct one is:
y' = ( 1- 3x^2 y^3 ) / ( 3x^3 y^2 -1)