# Implicit differentiation

• Oct 15th 2006, 04:38 PM
becky
Implicit differentiation
Find dy/dx by implicit differentiation.

(3xy^2 + 1)^4 = 2x - 3y

4(3xy^2 + 1)^3 (3xy^2 + 1) = 2 -3

(3xy^2 + 1) +3 = 2-4(3xy^2 +1)^3

dy/dx = 2 - 4(3xy^2 +1)^3/(3xy^2 +1) +3

Am I getting any closer to understanding this?
• Oct 15th 2006, 05:22 PM
Jameson
You started out ok, but your second line is incorrect.

This part is right 4(3xy^2+1)^3. Now you must multiply by the derivative of the inside function.

So d/dx (3xy^2+1)= 6xy(dy/dx)+3y^2.

Putting the LHS together it's 4(3xy^2+1)^3 * (6xy(dy/dx)+3y^2)

The RHS of your work is correct.
• Oct 15th 2006, 08:04 PM
Soroban
Hello, Becky!

I don't see dy/dx or y' in your work.
Then it suddenly appears in your last step.

Quote:

Find dy/dx by implicit differentiation: .(3xy² + 1)^4 .= .2x - 3y

. . . . . . 4(3xy² + 1)³(6xyy' + 3y²) . = . 2 - 3y'

. . . . . . 24xy(3xy² + 1)3y' + 12y²(3xy² + 1)³ . = . 2 - 3y'

. . . . . . 24xy(3xy² + 1)3y' + 3y' . = . 2 - 12y²(3xy² + 1)³

Factor: .3[8xy(3xy(3xy² + 1)³ + 1]y' . = . 2[1 - 6y²(3xy² + 1)³]

. . . . . . . . . . . . . . .2[1 - 6y²(3xy² + 1)³]
Therefore: . y' . = . ---------------------------
. . . . . . . . . . . . . . 3[8xy(3xy² + 1)³ + 1]

• Oct 16th 2006, 08:55 PM
earboth
Quote:

Originally Posted by becky
Find dy/dx by implicit differentiation.

(3xy^2 + 1)^4 = 2x - 3y...

Hi,

I've attached an image to show you what I've calculated: