Question

i have some questions about implicit differentiation

Answer 1

\[xy(2x+3y)=2\]

Answer 2

ok my question is do i need to set up the multiplication derivative equation for both xy(2x+3y) as well as xy?

Answer 3

You could distribute there!

Answer 4

ok so i should distribute first makes sense

Answer 5

Well it will make your life a lot easier

Answer 6

so know i have \[2x^2y+2xY^2=2 \] so know i am going to have to differentiate using the product equation for both 2x^2y and 2xy^2 right?

Answer 7

sorry that should be 3xy^2

Answer 8

Yes since you have a product for both terms you use the product rule for both terms

Answer 9

You might consider some of these useful for your problem:
\[(fg)'=f'g+fg' ; (cfg)'=c(fg)'; (y^n)'=ny^{n-1}y' ; (x^n)'=nx^{n-1}\]

Answer 10

\[(3xy^2)'=3(xy^2)' \text{ by constant multiple rule}\]
\[=3 \cdot [(x)'y^2+x(y^2)'] \text{ by product rule}\]
So we know (x)'=1
So we have:
\[=3[1y^2+x(2y^{2-1}y')] \text{ by chain rule}\]
So finally we can say all nice like that
\[(3xy^2)'=3[y^2+2xyy']\]

Answer 11

\[(3xy^2)'=3y^2+6xyy'\]

Answer 12

Do you see how I applied the rules?

Answer 13

You try the other product and I will check k?

Answer 14

ok

Answer 15

This one:
\[(3x^2y)'\]
And remember that y=f(x) :means y is a function of x

Answer 16

ok i got 4xy+2x^2y^prime

Answer 17

and i used (2x^2y)prime

Answer 18

Ok!
\[(2x^2y)'=2(x^2y)'=2(2xy+x^2y')=4xy+2x^2y'\]
Beautiful! :)

Answer 19

sweet! thanks a ton!

Answer 20

np! :)