Question

A question about implicit differentiation..
x^3 + y^3 - 9xy = 0

Answer 1

\[\large \frac{ d }{ dx } x^3 + \frac{ d }{ dx }y^3 - \frac{ d }{ dx } 9xy = \frac{ d }{ dx }0\]
I know how to derive the x^3 and y^3 and 0 ofc..
but I don't understand what I am suppose to with the -9xy..

Answer 2

use product rule d(xy)= x.dy+y.dx

Answer 3

Oh. That makes sense.

Answer 4

so
d/dx (-9xy) = -9y + -9x(1)?..

Answer 5

no, d/dx(-9xy) = -9x (dy/dx) - 9y

Answer 6

actually is d(x)/dx (-9y) + d(y)/dx(-9x)

Answer 7

then become 1(-9y)+dy/dx(-9x)

Answer 8

then become -9y-dy/dx(9x)

Answer 9

Why is the dy/dx still there though? ..

Answer 10

because chain rule state tht d/dx(xy) = dx/dx (y)+dy/dx(y)

Answer 11

sry is product rule... XD

Answer 12

so when ever u hv 2 variable, u will need to apply this product rule

Answer 13

it's fine lol. i'm still a bit lost though.
According to my book it becomes..
\[\large 3x^2 + 3y^2 \frac{ d }{ dx } - 9 (x \frac{ dy }{ dx } + y \frac{ dx }{ dx }) = 0\]
Couldn't x dy/dx be written as xy' ?

Answer 14

x (dy/dx) and xy' is the same things, but 2nd term in not correct, when u hv d/dx(3y^3) u need to take care, i will become like this, dy/dx*d/dy(3y^3)

Answer 15

sry is dy/dx*d/dy(y^3), then (dy/dx)(3y^2)

Answer 16

remember this, \[\frac{ d }{ dx }=\frac{ dy }{ dx }\times \frac{ d }{ dy }\]

Answer 17

see is a chain rule, when u cancel out the dy u will get back d/dx

Answer 18

leibniz notation always confuses me lol.

Answer 19

erm, i get use to this notation... ><, do u clear abt this?? or still blur?

Answer 20

when u r dealing with implicit differentiation, u go to term by term, so tht u will be easier, for checking and solving.

Answer 21

x^3 + y^3 - 9xy = 0
first term d/dx(x^3)=3x^2
second term d/dx(y^3)=(dy/dx)(d/dy)(y^3)=(dy/dx)3y^2
third term d/dx(-9xy)=-9(d/dx(xy))=-9(x(dy/dx)+y(dx/dx))=-9(x(dy/dx)+y)
overall, 3x^2 + (dy/dx)3y^2 - -9(x(dy/dx)+y) = 0

Answer 22

I think I got it .
\[\large \frac{ d }{ dx } (x^3+y^3-9xy = 0)\]
\[\large 3x^2 + 3y^2y' - 9xy' - 9y = 0\]
Group the y' together and get terms on one side.
\[\large y'(3y^2 - 9x) = -3x^2 + 9x\]
Divide.
\[\large y' = \frac{ 3(-x^2+3x) }{ 3(y^2-3x) } = \frac{ -x^2+3x }{ y^2-3x }\]

Answer 23

yup, urs correct, my equation hv some error on 3rd term is 1 minus sign only...

Answer 24

Oops, I made a mistake i accidentally made the -9y into 9x instead of 9y*

Answer 25

\[\large y' = \frac{ 3y - x^2 }{ y^2 - 3x }\]
there we go.

Answer 26

yup =)

Answer 27

Thanks.