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?

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- Oct 15th 2006, 04:38 PMbeckyImplicit 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 PMJameson
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 PMSoroban
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(6xy**y'**+ 3y²) . = . 2 - 3**y'**

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

. . . . . . 24xy(3xy² + 1)3**y'**+ 3**y'**. = . 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 PMearboth