# Implicit Differentiation

• Mar 7th 2007, 08:17 PM
drain
Implicit Differentiation
I got past the first step, but I am having trouble solving for d/dx

(x^(2) + y^(2))^(3) = ax^(2)y a is constant

for the first step i got

3(x^(2) + y^(2))^(2) * (2x + 2y (d/dx)) = ax^(2) (d/dx) + 2axy

after that I'm stuck...:(
• Mar 7th 2007, 08:37 PM
Jhevon
Quote:

Originally Posted by drain
I got past the first step, but I am having trouble solving for d/dx

(x^(2) + y^(2))^(3) = ax^(2)y a is constant

for the first step i got

3(x^(2) + y^(2))^(2) * (2x + 2y (d/dx)) = ax^(2) (d/dx) + 2axy

after that I'm stuck...:(

dont you mean you want to solve for dy/dx?

(x^(2) + y^(2))^(3) = ax^(2)y ..........its painful, but i think it'll help to expand these brackets.

=> x^6 + 3x^4y^2 + 3x^2y^4 + y^6 = ax^2y

now we go through and differentiate implicitly. we add dy/dx to a function when we differentiate with respect to y, we add dx/dx when we differentiate with respect to x, but dx/dx cancels into 1, so we actually don't write it. also note that we need the product rule when we have x * y in any form. so here goes.

=> 6x^5 + 12x^3y^2 + 6x^4y dy/dx + 6xy^4 + 12x^2y^3 dy/dx + 6y^5 dy/dx = 2axy + ax^2 dy/dx ...........now we group everything with dy/dx on one side so we can factor them out.

=> 6x^4y dy/dx + 12x^2y^3 dy/dx + 6y^5 dy/dx - ax^2 dy/dx = 2axy - 6x^5 - 12x^3y^2 - 6xy^4

=> dy/dx(6x^4y + 12x^2y^3 + 6y^5 - ax^2) = 2axy - 6x^5 - 12x^3y^2 - 6xy^4
=> dy/dx = (2axy - 6x^5 - 12x^3y^2 - 6xy^4)/(6x^4y + 12x^2y^3 + 6y^5 - ax^2)

check if i expanded the cubic correctly
• Mar 7th 2007, 09:28 PM
Soroban
Hello, drain!

Quote:

I got past the first step, but I am having trouble solving for d/dx

. . (x² + y²)³ .= .ax²y . . a is constant

For the first step i got:
. . 3(x² + y²)²(2x + 2yy') .= .ax²y' + 2axy . Good!

Expand the left side (partially): .6x(x² + y²)² + 6y(x² + y²)²y' .= .ax²y' + 2axy

Rearrange terms: .6y(x² + y²)²y' - ax²y' .= .2axy - 6x(x² + y²)²

Factor: . [6y(x² + y²)² - ax²]y' .= .2x[ay - 3(x² + y²)²]

. . . . . . . . . . . . . . 2x[ay - 3(x² + y²)²]
Therefore: . y' . = . ------------------------
. . . . . . . . . . . . . . .6y(x² + y²)² - ax²

• Mar 7th 2007, 11:27 PM
drain
Thanks alot! That one was realllly tricky... and the way that other user did it... I appreciate the help but it confused the heck out of me! :)